That the given differential equation has a regular singular point at x = 0 is and determine two solutions for x > 0 by using the given data.

Question

Want to see more full solutions like this?

Answer 1

Question

Chapter 5.7, Problem 1P

To determine

To show: That the given differential equation has a regular singular point at $x=0$ is and determine two solutions for $x>0$ by using the given data.

Expert Solution & Answer

Answer to Problem 1P

The first solution of the differential equation is $y1(x)=∑n=0∞(−1)nxnn!(n+1)!_$ and second solution is $y2(x)=−y1(x)lnx+x−1[1+∑n=1∞(−1)n+1Hn+Hn−1n!(n−1)!xn]_$ .

Explanation of Solution

Theorem used:

Let $r1$ and $r2$ be the roots of the indicial equation $F(r)=r(r−1)+p0r+q0$ with $r1≥r2$ .

If $r1$ and $r2$ are real, then in either the interval $−ρ<x<0$ or the interval $0<x<ρ$ , there exists a solution of the form $y1(x)=|x|r1[1+∑n=1∞an(r1)xn]$ where $an(r1)$ are the given by recurrence relation with $a0=1$ and $r=r1$ .

If $r1−r2$ is not zero or a positive integer, then in either the interval $−ρ<x<0$ or the interval $0<x<ρ$ , there exists a second solution of the form $y2(x)=|x|r2[1+∑n=1∞an(r2)xn]$ where $an(r2)$ are also determined by the recurrence relation with $a0=1$ and $r=r2$ .

Further the power series converges at least for $|x|<ρ$ .

If $r1=r2$ , then the second solution is $y2(x)=y1(x)ln|x|+|x|r1[1+∑n=1∞bn(r1)xn]$ .

If $r1−r2=N$ , positive integer, then $y2(x)=ay1(x)ln|x|+|x|r2[1+∑n=1∞cn(r2)xn]$ .

Calculation:

The given differential equation is given as $x2y"+2xy'+xy=0$ .

Compare the equation $x2y"+2xy'+xy=0$ with $P(x)y"+Q(x)y'+R(x)y=0$ and note that $P(x)=x2$ , $Q(x)=2x$ and $R(x)=x$ .

Note that the singular points occur when $P(x0)=0$ .

Thus, the singular point of the given equation is $x0=0$ .

If $x0=0$ is a regular singular point, check whether the following limits are finite.

$p0=limx→0xQ(x)P(x)$ and $q0=limx→0x2R(x)P(x)$

For $x0=0$ , obtain the value of $p0$ as follows.

$p0=limx→0x2xx2=2$

Similarly, for $x0=0$ , obtain the value of $q0$ as follows.

$q0=limx→0x2x2x2=0$ .

Since both are finite, $x0=0$ is a regular singular point of the given differential equation.

Since $p0=2$ and $q0=0$ , the corresponding indicial equation is as follows:

$F(r)=r(r−1)+p0r+q0$

Substitute the value of $p0=2$ and $q0=0$ in $F(r)=r(r−1)+p0r+q0=0$ .

$F(r)=r(r−1)+2r+0=r2−r+2r=r2+r$

Let $F(r)=0$ in the above equation it becomes as follows:

$r2+r=0r(r+1)=0$

Therefore, the roots of the equation $r2−r+2r$ is $r=0,r=−1$ .

Hence, $x0$ is a regular singular point.

From the above Theorem, the first solution is given as $y1(x)=|x|r1[1+∑n=1∞an(r1)xn]$ .

Substitute the value of $r=0$ in $y1(x)=|x|r1[1+∑n=1∞an(r1)xn]$ .

$y1(x)=1+∑n=1∞an(0)xn$

Differentiate the equation $y1(x)=1+∑n=1∞an(0)xn$ .

$y1'(x)=∑n=1∞(n+1)an+1(0)xn$

The coefficients $an(0)$ , apart from $a0(0)=1$ are determined by the recurrence relation $F(r+n)an+∑k=1n−1ak[(r+k)pn−k+qn−k]=0$ in case of $r=r1$ .

$n(n+1)an(0)+∑k=1n−1ak(0)[kpn−k+qn−k]=0$ , where $∑n=0∞pnxn=xQ(x)P(x)=2$ and $∑n=0∞qnxn=x2R(x)P(x)=x$ .

Hence, the value of $p0=2,pn=0,∀n∈N$ and $q1=1,qn=0,∀n∈N0$ .

Factoring out $an(0)$ from the equation $n(n+1)an(0)+∑k=1n−1ak(0)[kpn−k+qn−k]=0$ is as follows:

$an(0)=−∑k=0n−1ak(0)[kpn−k+qn−k]n(n+1)$ with first couple of elements is as follows:

$a1(0)=−a0(0)[0+q1]1(1+1)=−11⋅2a2(0)=11⋅22⋅3a3(0)=−11⋅22⋅32⋅4$

And the values continues as such.

Hence, the first solution is $y1(x)=∑n=0∞(−1)nxnn!(n+1)!_$ .

According to the same theorem, since $r1−r2=1$ , the second solution is given by $y2(x)=ay1(x)ln|x|+|x|r2[1+∑n=1∞cn(r2)xn]$ .

$y2(x)=ay1(x)ln|x|+x−1[1+∑n=1∞cn(−1)xn]=ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn$

Where $a=limr→r2(r−r2)aN(r),N=r−r2$ since $N=1$ .

$aN(r)=a1(r)=−∑k=0N−1ak(r)[(r+k)pN−k+qN−k](r+N)(r+N+1)$

Substitute the limits in the above equation it becomes as follows:

$a1(r)=−a0(r)[(r+0)p1+q1](r+2)(r+1)=−[0+1](r+2)(r+1)=−1(r+2)(r+1)$

Thus, substitute the value of $a=limr→−1(r+1)−1(r+2)(r+1)=−1$ .

To calculate the value of $cn(0)$ substitute the value of $y2(x)=ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn$ into the initial differential equation.

First differentiate the equation $y2(x)=ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn$ .

$y2'(x)=ay1'(x)lnx+ay1(x)x−x−2+∑n=1∞ncn+1(−1)xn−1$

Again, differentiate the equation $y2'(x)=ay1'(x)lnx+ay1(x)x−x−2+∑n=1∞ncn+1(−1)xn−1$ .

$y2''(x)=ay1''(x)lnx+2ay1'(x)x−ay1(x)x+2x−3+∑n=1∞n(n−1)cn+1(−1)xn−2$

Now, the given differential equation becomes as follows:

$x2[ay1''(x)lnx+2ay1'(x)x−ay1(x)x+2x−3+∑n=1∞n(n−1)cn+1(−1)xn−2]+2x[ay1'(x)lnx+ay1(x)x−x−2+∑n=1∞ncn+1(−1)xn−1]+x[ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn]=0[2c2(−1)+c1(−1)−3a1(0)]x+∑n=1∞[−(2n+1)an(0)+n(n+1)]cn+1(−1)cn(−1)xn=0$

Equating the coefficient on both sides in the above equation.

${2c2(−1)+c1(−1)−3a1(0)=0−(2n+1)an(0)+n(n+1)cn+1(−1)cn(−1)=0$

Let $c2(−1)=1$ in the above equation this implies as follows:

$c2(−1)=3a1(0)−c1(−1)2=−H2+H12!c3(−1)=H3+H22⋅3$

$cn(−1)=(−1)n+1Hn+Hn−1n!(n−1)!$

Hence, the second solution is $y2(x)=−y1(x)lnx+x−1+[1+∑n=1∞(−1)n+1Hn+Hn−1n!(n−1)!xn]_$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

B\ Prove that if T is a spanning tree of G which contains e, then Te Is a spanning tree of G * e.

9 Q/ Let G be agraph with n vertices, then G has at least two vertices which are not cut vertices.

Find the first four nonzero terms in a power series expansion about x=0 for a general solution to the given differential equation w''-14x^2w'+w=0

Answer 2

Question

Chapter 5.7, Problem 1P

To determine

To show: That the given differential equation has a regular singular point at $x=0$ is and determine two solutions for $x>0$ by using the given data.

Expert Solution & Answer

Answer to Problem 1P

The first solution of the differential equation is $y1(x)=∑n=0∞(−1)nxnn!(n+1)!_$ and second solution is $y2(x)=−y1(x)lnx+x−1[1+∑n=1∞(−1)n+1Hn+Hn−1n!(n−1)!xn]_$ .

Explanation of Solution

Theorem used:

Let $r1$ and $r2$ be the roots of the indicial equation $F(r)=r(r−1)+p0r+q0$ with $r1≥r2$ .

If $r1$ and $r2$ are real, then in either the interval $−ρ<x<0$ or the interval $0<x<ρ$ , there exists a solution of the form $y1(x)=|x|r1[1+∑n=1∞an(r1)xn]$ where $an(r1)$ are the given by recurrence relation with $a0=1$ and $r=r1$ .

If $r1−r2$ is not zero or a positive integer, then in either the interval $−ρ<x<0$ or the interval $0<x<ρ$ , there exists a second solution of the form $y2(x)=|x|r2[1+∑n=1∞an(r2)xn]$ where $an(r2)$ are also determined by the recurrence relation with $a0=1$ and $r=r2$ .

Further the power series converges at least for $|x|<ρ$ .

If $r1=r2$ , then the second solution is $y2(x)=y1(x)ln|x|+|x|r1[1+∑n=1∞bn(r1)xn]$ .

If $r1−r2=N$ , positive integer, then $y2(x)=ay1(x)ln|x|+|x|r2[1+∑n=1∞cn(r2)xn]$ .

Calculation:

The given differential equation is given as $x2y"+2xy'+xy=0$ .

Compare the equation $x2y"+2xy'+xy=0$ with $P(x)y"+Q(x)y'+R(x)y=0$ and note that $P(x)=x2$ , $Q(x)=2x$ and $R(x)=x$ .

Note that the singular points occur when $P(x0)=0$ .

Thus, the singular point of the given equation is $x0=0$ .

If $x0=0$ is a regular singular point, check whether the following limits are finite.

$p0=limx→0xQ(x)P(x)$ and $q0=limx→0x2R(x)P(x)$

For $x0=0$ , obtain the value of $p0$ as follows.

$p0=limx→0x2xx2=2$

Similarly, for $x0=0$ , obtain the value of $q0$ as follows.

$q0=limx→0x2x2x2=0$ .

Since both are finite, $x0=0$ is a regular singular point of the given differential equation.

Since $p0=2$ and $q0=0$ , the corresponding indicial equation is as follows:

$F(r)=r(r−1)+p0r+q0$

Substitute the value of $p0=2$ and $q0=0$ in $F(r)=r(r−1)+p0r+q0=0$ .

$F(r)=r(r−1)+2r+0=r2−r+2r=r2+r$

Let $F(r)=0$ in the above equation it becomes as follows:

$r2+r=0r(r+1)=0$

Therefore, the roots of the equation $r2−r+2r$ is $r=0,r=−1$ .

Hence, $x0$ is a regular singular point.

From the above Theorem, the first solution is given as $y1(x)=|x|r1[1+∑n=1∞an(r1)xn]$ .

Substitute the value of $r=0$ in $y1(x)=|x|r1[1+∑n=1∞an(r1)xn]$ .

$y1(x)=1+∑n=1∞an(0)xn$

Differentiate the equation $y1(x)=1+∑n=1∞an(0)xn$ .

$y1'(x)=∑n=1∞(n+1)an+1(0)xn$

The coefficients $an(0)$ , apart from $a0(0)=1$ are determined by the recurrence relation $F(r+n)an+∑k=1n−1ak[(r+k)pn−k+qn−k]=0$ in case of $r=r1$ .

$n(n+1)an(0)+∑k=1n−1ak(0)[kpn−k+qn−k]=0$ , where $∑n=0∞pnxn=xQ(x)P(x)=2$ and $∑n=0∞qnxn=x2R(x)P(x)=x$ .

Hence, the value of $p0=2,pn=0,∀n∈N$ and $q1=1,qn=0,∀n∈N0$ .

Factoring out $an(0)$ from the equation $n(n+1)an(0)+∑k=1n−1ak(0)[kpn−k+qn−k]=0$ is as follows:

$an(0)=−∑k=0n−1ak(0)[kpn−k+qn−k]n(n+1)$ with first couple of elements is as follows:

$a1(0)=−a0(0)[0+q1]1(1+1)=−11⋅2a2(0)=11⋅22⋅3a3(0)=−11⋅22⋅32⋅4$

And the values continues as such.

Hence, the first solution is $y1(x)=∑n=0∞(−1)nxnn!(n+1)!_$ .

According to the same theorem, since $r1−r2=1$ , the second solution is given by $y2(x)=ay1(x)ln|x|+|x|r2[1+∑n=1∞cn(r2)xn]$ .

$y2(x)=ay1(x)ln|x|+x−1[1+∑n=1∞cn(−1)xn]=ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn$

Where $a=limr→r2(r−r2)aN(r),N=r−r2$ since $N=1$ .

$aN(r)=a1(r)=−∑k=0N−1ak(r)[(r+k)pN−k+qN−k](r+N)(r+N+1)$

Substitute the limits in the above equation it becomes as follows:

$a1(r)=−a0(r)[(r+0)p1+q1](r+2)(r+1)=−[0+1](r+2)(r+1)=−1(r+2)(r+1)$

Thus, substitute the value of $a=limr→−1(r+1)−1(r+2)(r+1)=−1$ .

To calculate the value of $cn(0)$ substitute the value of $y2(x)=ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn$ into the initial differential equation.

First differentiate the equation $y2(x)=ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn$ .

$y2'(x)=ay1'(x)lnx+ay1(x)x−x−2+∑n=1∞ncn+1(−1)xn−1$

Again, differentiate the equation $y2'(x)=ay1'(x)lnx+ay1(x)x−x−2+∑n=1∞ncn+1(−1)xn−1$ .

$y2''(x)=ay1''(x)lnx+2ay1'(x)x−ay1(x)x+2x−3+∑n=1∞n(n−1)cn+1(−1)xn−2$

Now, the given differential equation becomes as follows:

$x2[ay1''(x)lnx+2ay1'(x)x−ay1(x)x+2x−3+∑n=1∞n(n−1)cn+1(−1)xn−2]+2x[ay1'(x)lnx+ay1(x)x−x−2+∑n=1∞ncn+1(−1)xn−1]+x[ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn]=0[2c2(−1)+c1(−1)−3a1(0)]x+∑n=1∞[−(2n+1)an(0)+n(n+1)]cn+1(−1)cn(−1)xn=0$

Equating the coefficient on both sides in the above equation.

${2c2(−1)+c1(−1)−3a1(0)=0−(2n+1)an(0)+n(n+1)cn+1(−1)cn(−1)=0$

Let $c2(−1)=1$ in the above equation this implies as follows:

$c2(−1)=3a1(0)−c1(−1)2=−H2+H12!c3(−1)=H3+H22⋅3$

$cn(−1)=(−1)n+1Hn+Hn−1n!(n−1)!$

Hence, the second solution is $y2(x)=−y1(x)lnx+x−1+[1+∑n=1∞(−1)n+1Hn+Hn−1n!(n−1)!xn]_$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Chapter 5.7, Problem 1P

To determine

To show: That the given differential equation has a regular singular point at $x=0$ is and determine two solutions for $x>0$ by using the given data.

Expert Solution & Answer

Answer to Problem 1P

The first solution of the differential equation is $y1(x)=∑n=0∞(−1)nxnn!(n+1)!_$ and second solution is $y2(x)=−y1(x)lnx+x−1[1+∑n=1∞(−1)n+1Hn+Hn−1n!(n−1)!xn]_$ .

Explanation of Solution

Theorem used:

Let $r1$ and $r2$ be the roots of the indicial equation $F(r)=r(r−1)+p0r+q0$ with $r1≥r2$ .

If $r1$ and $r2$ are real, then in either the interval $−ρ<x<0$ or the interval $0<x<ρ$ , there exists a solution of the form $y1(x)=|x|r1[1+∑n=1∞an(r1)xn]$ where $an(r1)$ are the given by recurrence relation with $a0=1$ and $r=r1$ .

If $r1−r2$ is not zero or a positive integer, then in either the interval $−ρ<x<0$ or the interval $0<x<ρ$ , there exists a second solution of the form $y2(x)=|x|r2[1+∑n=1∞an(r2)xn]$ where $an(r2)$ are also determined by the recurrence relation with $a0=1$ and $r=r2$ .

Further the power series converges at least for $|x|<ρ$ .

If $r1=r2$ , then the second solution is $y2(x)=y1(x)ln|x|+|x|r1[1+∑n=1∞bn(r1)xn]$ .

If $r1−r2=N$ , positive integer, then $y2(x)=ay1(x)ln|x|+|x|r2[1+∑n=1∞cn(r2)xn]$ .

Calculation:

The given differential equation is given as $x2y"+2xy'+xy=0$ .

Compare the equation $x2y"+2xy'+xy=0$ with $P(x)y"+Q(x)y'+R(x)y=0$ and note that $P(x)=x2$ , $Q(x)=2x$ and $R(x)=x$ .

Note that the singular points occur when $P(x0)=0$ .

Thus, the singular point of the given equation is $x0=0$ .

If $x0=0$ is a regular singular point, check whether the following limits are finite.

$p0=limx→0xQ(x)P(x)$ and $q0=limx→0x2R(x)P(x)$

For $x0=0$ , obtain the value of $p0$ as follows.

$p0=limx→0x2xx2=2$

Similarly, for $x0=0$ , obtain the value of $q0$ as follows.

$q0=limx→0x2x2x2=0$ .

Since both are finite, $x0=0$ is a regular singular point of the given differential equation.

Since $p0=2$ and $q0=0$ , the corresponding indicial equation is as follows:

$F(r)=r(r−1)+p0r+q0$

Substitute the value of $p0=2$ and $q0=0$ in $F(r)=r(r−1)+p0r+q0=0$ .

$F(r)=r(r−1)+2r+0=r2−r+2r=r2+r$

Let $F(r)=0$ in the above equation it becomes as follows:

$r2+r=0r(r+1)=0$

Therefore, the roots of the equation $r2−r+2r$ is $r=0,r=−1$ .

Hence, $x0$ is a regular singular point.

From the above Theorem, the first solution is given as $y1(x)=|x|r1[1+∑n=1∞an(r1)xn]$ .

Substitute the value of $r=0$ in $y1(x)=|x|r1[1+∑n=1∞an(r1)xn]$ .

$y1(x)=1+∑n=1∞an(0)xn$

Differentiate the equation $y1(x)=1+∑n=1∞an(0)xn$ .

$y1'(x)=∑n=1∞(n+1)an+1(0)xn$

The coefficients $an(0)$ , apart from $a0(0)=1$ are determined by the recurrence relation $F(r+n)an+∑k=1n−1ak[(r+k)pn−k+qn−k]=0$ in case of $r=r1$ .

$n(n+1)an(0)+∑k=1n−1ak(0)[kpn−k+qn−k]=0$ , where $∑n=0∞pnxn=xQ(x)P(x)=2$ and $∑n=0∞qnxn=x2R(x)P(x)=x$ .

Hence, the value of $p0=2,pn=0,∀n∈N$ and $q1=1,qn=0,∀n∈N0$ .

Factoring out $an(0)$ from the equation $n(n+1)an(0)+∑k=1n−1ak(0)[kpn−k+qn−k]=0$ is as follows:

$an(0)=−∑k=0n−1ak(0)[kpn−k+qn−k]n(n+1)$ with first couple of elements is as follows:

$a1(0)=−a0(0)[0+q1]1(1+1)=−11⋅2a2(0)=11⋅22⋅3a3(0)=−11⋅22⋅32⋅4$

And the values continues as such.

Hence, the first solution is $y1(x)=∑n=0∞(−1)nxnn!(n+1)!_$ .

According to the same theorem, since $r1−r2=1$ , the second solution is given by $y2(x)=ay1(x)ln|x|+|x|r2[1+∑n=1∞cn(r2)xn]$ .

$y2(x)=ay1(x)ln|x|+x−1[1+∑n=1∞cn(−1)xn]=ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn$

Where $a=limr→r2(r−r2)aN(r),N=r−r2$ since $N=1$ .

$aN(r)=a1(r)=−∑k=0N−1ak(r)[(r+k)pN−k+qN−k](r+N)(r+N+1)$

Substitute the limits in the above equation it becomes as follows:

$a1(r)=−a0(r)[(r+0)p1+q1](r+2)(r+1)=−[0+1](r+2)(r+1)=−1(r+2)(r+1)$

Thus, substitute the value of $a=limr→−1(r+1)−1(r+2)(r+1)=−1$ .

To calculate the value of $cn(0)$ substitute the value of $y2(x)=ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn$ into the initial differential equation.

First differentiate the equation $y2(x)=ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn$ .

$y2'(x)=ay1'(x)lnx+ay1(x)x−x−2+∑n=1∞ncn+1(−1)xn−1$

Again, differentiate the equation $y2'(x)=ay1'(x)lnx+ay1(x)x−x−2+∑n=1∞ncn+1(−1)xn−1$ .

$y2''(x)=ay1''(x)lnx+2ay1'(x)x−ay1(x)x+2x−3+∑n=1∞n(n−1)cn+1(−1)xn−2$

Now, the given differential equation becomes as follows:

$x2[ay1''(x)lnx+2ay1'(x)x−ay1(x)x+2x−3+∑n=1∞n(n−1)cn+1(−1)xn−2]+2x[ay1'(x)lnx+ay1(x)x−x−2+∑n=1∞ncn+1(−1)xn−1]+x[ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn]=0[2c2(−1)+c1(−1)−3a1(0)]x+∑n=1∞[−(2n+1)an(0)+n(n+1)]cn+1(−1)cn(−1)xn=0$

Equating the coefficient on both sides in the above equation.

${2c2(−1)+c1(−1)−3a1(0)=0−(2n+1)an(0)+n(n+1)cn+1(−1)cn(−1)=0$

Let $c2(−1)=1$ in the above equation this implies as follows:

$c2(−1)=3a1(0)−c1(−1)2=−H2+H12!c3(−1)=H3+H22⋅3$

$cn(−1)=(−1)n+1Hn+Hn−1n!(n−1)!$

Hence, the second solution is $y2(x)=−y1(x)lnx+x−1+[1+∑n=1∞(−1)n+1Hn+Hn−1n!(n−1)!xn]_$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Chapter 5.7, Problem 1P

To determine

To show: That the given differential equation has a regular singular point at $x=0$ is and determine two solutions for $x>0$ by using the given data.

Expert Solution & Answer

Answer to Problem 1P

The first solution of the differential equation is $y1(x)=∑n=0∞(−1)nxnn!(n+1)!_$ and second solution is $y2(x)=−y1(x)lnx+x−1[1+∑n=1∞(−1)n+1Hn+Hn−1n!(n−1)!xn]_$ .

Explanation of Solution

Theorem used:

Let $r1$ and $r2$ be the roots of the indicial equation $F(r)=r(r−1)+p0r+q0$ with $r1≥r2$ .

If $r1$ and $r2$ are real, then in either the interval $−ρ<x<0$ or the interval $0<x<ρ$ , there exists a solution of the form $y1(x)=|x|r1[1+∑n=1∞an(r1)xn]$ where $an(r1)$ are the given by recurrence relation with $a0=1$ and $r=r1$ .

If $r1−r2$ is not zero or a positive integer, then in either the interval $−ρ<x<0$ or the interval $0<x<ρ$ , there exists a second solution of the form $y2(x)=|x|r2[1+∑n=1∞an(r2)xn]$ where $an(r2)$ are also determined by the recurrence relation with $a0=1$ and $r=r2$ .

Further the power series converges at least for $|x|<ρ$ .

If $r1=r2$ , then the second solution is $y2(x)=y1(x)ln|x|+|x|r1[1+∑n=1∞bn(r1)xn]$ .

If $r1−r2=N$ , positive integer, then $y2(x)=ay1(x)ln|x|+|x|r2[1+∑n=1∞cn(r2)xn]$ .

Calculation:

The given differential equation is given as $x2y"+2xy'+xy=0$ .

Compare the equation $x2y"+2xy'+xy=0$ with $P(x)y"+Q(x)y'+R(x)y=0$ and note that $P(x)=x2$ , $Q(x)=2x$ and $R(x)=x$ .

Note that the singular points occur when $P(x0)=0$ .

Thus, the singular point of the given equation is $x0=0$ .

If $x0=0$ is a regular singular point, check whether the following limits are finite.

$p0=limx→0xQ(x)P(x)$ and $q0=limx→0x2R(x)P(x)$

For $x0=0$ , obtain the value of $p0$ as follows.

$p0=limx→0x2xx2=2$

Similarly, for $x0=0$ , obtain the value of $q0$ as follows.

$q0=limx→0x2x2x2=0$ .

Since both are finite, $x0=0$ is a regular singular point of the given differential equation.

Since $p0=2$ and $q0=0$ , the corresponding indicial equation is as follows:

$F(r)=r(r−1)+p0r+q0$

Substitute the value of $p0=2$ and $q0=0$ in $F(r)=r(r−1)+p0r+q0=0$ .

$F(r)=r(r−1)+2r+0=r2−r+2r=r2+r$

Let $F(r)=0$ in the above equation it becomes as follows:

$r2+r=0r(r+1)=0$

Therefore, the roots of the equation $r2−r+2r$ is $r=0,r=−1$ .

Hence, $x0$ is a regular singular point.

From the above Theorem, the first solution is given as $y1(x)=|x|r1[1+∑n=1∞an(r1)xn]$ .

Substitute the value of $r=0$ in $y1(x)=|x|r1[1+∑n=1∞an(r1)xn]$ .

$y1(x)=1+∑n=1∞an(0)xn$

Differentiate the equation $y1(x)=1+∑n=1∞an(0)xn$ .

$y1'(x)=∑n=1∞(n+1)an+1(0)xn$

The coefficients $an(0)$ , apart from $a0(0)=1$ are determined by the recurrence relation $F(r+n)an+∑k=1n−1ak[(r+k)pn−k+qn−k]=0$ in case of $r=r1$ .

$n(n+1)an(0)+∑k=1n−1ak(0)[kpn−k+qn−k]=0$ , where $∑n=0∞pnxn=xQ(x)P(x)=2$ and $∑n=0∞qnxn=x2R(x)P(x)=x$ .

Hence, the value of $p0=2,pn=0,∀n∈N$ and $q1=1,qn=0,∀n∈N0$ .

Factoring out $an(0)$ from the equation $n(n+1)an(0)+∑k=1n−1ak(0)[kpn−k+qn−k]=0$ is as follows:

$an(0)=−∑k=0n−1ak(0)[kpn−k+qn−k]n(n+1)$ with first couple of elements is as follows:

$a1(0)=−a0(0)[0+q1]1(1+1)=−11⋅2a2(0)=11⋅22⋅3a3(0)=−11⋅22⋅32⋅4$

And the values continues as such.

Hence, the first solution is $y1(x)=∑n=0∞(−1)nxnn!(n+1)!_$ .

According to the same theorem, since $r1−r2=1$ , the second solution is given by $y2(x)=ay1(x)ln|x|+|x|r2[1+∑n=1∞cn(r2)xn]$ .

$y2(x)=ay1(x)ln|x|+x−1[1+∑n=1∞cn(−1)xn]=ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn$

Where $a=limr→r2(r−r2)aN(r),N=r−r2$ since $N=1$ .

$aN(r)=a1(r)=−∑k=0N−1ak(r)[(r+k)pN−k+qN−k](r+N)(r+N+1)$

Substitute the limits in the above equation it becomes as follows:

$a1(r)=−a0(r)[(r+0)p1+q1](r+2)(r+1)=−[0+1](r+2)(r+1)=−1(r+2)(r+1)$

Thus, substitute the value of $a=limr→−1(r+1)−1(r+2)(r+1)=−1$ .

To calculate the value of $cn(0)$ substitute the value of $y2(x)=ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn$ into the initial differential equation.

First differentiate the equation $y2(x)=ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn$ .

$y2'(x)=ay1'(x)lnx+ay1(x)x−x−2+∑n=1∞ncn+1(−1)xn−1$

Again, differentiate the equation $y2'(x)=ay1'(x)lnx+ay1(x)x−x−2+∑n=1∞ncn+1(−1)xn−1$ .

$y2''(x)=ay1''(x)lnx+2ay1'(x)x−ay1(x)x+2x−3+∑n=1∞n(n−1)cn+1(−1)xn−2$

Now, the given differential equation becomes as follows:

$x2[ay1''(x)lnx+2ay1'(x)x−ay1(x)x+2x−3+∑n=1∞n(n−1)cn+1(−1)xn−2]+2x[ay1'(x)lnx+ay1(x)x−x−2+∑n=1∞ncn+1(−1)xn−1]+x[ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn]=0[2c2(−1)+c1(−1)−3a1(0)]x+∑n=1∞[−(2n+1)an(0)+n(n+1)]cn+1(−1)cn(−1)xn=0$

Equating the coefficient on both sides in the above equation.

${2c2(−1)+c1(−1)−3a1(0)=0−(2n+1)an(0)+n(n+1)cn+1(−1)cn(−1)=0$

Let $c2(−1)=1$ in the above equation this implies as follows:

$c2(−1)=3a1(0)−c1(−1)2=−H2+H12!c3(−1)=H3+H22⋅3$

$cn(−1)=(−1)n+1Hn+Hn−1n!(n−1)!$

Hence, the second solution is $y2(x)=−y1(x)lnx+x−1+[1+∑n=1∞(−1)n+1Hn+Hn−1n!(n−1)!xn]_$ .

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 5.7, Problem 1P

To determine

To show: That the given differential equation has a regular singular point at $x=0$ is and determine two solutions for $x>0$ by using the given data.

Expert Solution & Answer

Answer to Problem 1P

The first solution of the differential equation is $y1(x)=∑n=0∞(−1)nxnn!(n+1)!_$ and second solution is $y2(x)=−y1(x)lnx+x−1[1+∑n=1∞(−1)n+1Hn+Hn−1n!(n−1)!xn]_$ .

Explanation of Solution

Theorem used:

Let $r1$ and $r2$ be the roots of the indicial equation $F(r)=r(r−1)+p0r+q0$ with $r1≥r2$ .

If $r1$ and $r2$ are real, then in either the interval $−ρ<x<0$ or the interval $0<x<ρ$ , there exists a solution of the form $y1(x)=|x|r1[1+∑n=1∞an(r1)xn]$ where $an(r1)$ are the given by recurrence relation with $a0=1$ and $r=r1$ .

If $r1−r2$ is not zero or a positive integer, then in either the interval $−ρ<x<0$ or the interval $0<x<ρ$ , there exists a second solution of the form $y2(x)=|x|r2[1+∑n=1∞an(r2)xn]$ where $an(r2)$ are also determined by the recurrence relation with $a0=1$ and $r=r2$ .

Further the power series converges at least for $|x|<ρ$ .

If $r1=r2$ , then the second solution is $y2(x)=y1(x)ln|x|+|x|r1[1+∑n=1∞bn(r1)xn]$ .

If $r1−r2=N$ , positive integer, then $y2(x)=ay1(x)ln|x|+|x|r2[1+∑n=1∞cn(r2)xn]$ .

Calculation:

The given differential equation is given as $x2y"+2xy'+xy=0$ .

Compare the equation $x2y"+2xy'+xy=0$ with $P(x)y"+Q(x)y'+R(x)y=0$ and note that $P(x)=x2$ , $Q(x)=2x$ and $R(x)=x$ .

Note that the singular points occur when $P(x0)=0$ .

Thus, the singular point of the given equation is $x0=0$ .

If $x0=0$ is a regular singular point, check whether the following limits are finite.

$p0=limx→0xQ(x)P(x)$ and $q0=limx→0x2R(x)P(x)$

For $x0=0$ , obtain the value of $p0$ as follows.

$p0=limx→0x2xx2=2$

Similarly, for $x0=0$ , obtain the value of $q0$ as follows.

$q0=limx→0x2x2x2=0$ .

Since both are finite, $x0=0$ is a regular singular point of the given differential equation.

Since $p0=2$ and $q0=0$ , the corresponding indicial equation is as follows:

$F(r)=r(r−1)+p0r+q0$

Substitute the value of $p0=2$ and $q0=0$ in $F(r)=r(r−1)+p0r+q0=0$ .

$F(r)=r(r−1)+2r+0=r2−r+2r=r2+r$

Let $F(r)=0$ in the above equation it becomes as follows:

$r2+r=0r(r+1)=0$

Therefore, the roots of the equation $r2−r+2r$ is $r=0,r=−1$ .

Hence, $x0$ is a regular singular point.

From the above Theorem, the first solution is given as $y1(x)=|x|r1[1+∑n=1∞an(r1)xn]$ .

Substitute the value of $r=0$ in $y1(x)=|x|r1[1+∑n=1∞an(r1)xn]$ .

$y1(x)=1+∑n=1∞an(0)xn$

Differentiate the equation $y1(x)=1+∑n=1∞an(0)xn$ .

$y1'(x)=∑n=1∞(n+1)an+1(0)xn$

The coefficients $an(0)$ , apart from $a0(0)=1$ are determined by the recurrence relation $F(r+n)an+∑k=1n−1ak[(r+k)pn−k+qn−k]=0$ in case of $r=r1$ .

$n(n+1)an(0)+∑k=1n−1ak(0)[kpn−k+qn−k]=0$ , where $∑n=0∞pnxn=xQ(x)P(x)=2$ and $∑n=0∞qnxn=x2R(x)P(x)=x$ .

Hence, the value of $p0=2,pn=0,∀n∈N$ and $q1=1,qn=0,∀n∈N0$ .

Factoring out $an(0)$ from the equation $n(n+1)an(0)+∑k=1n−1ak(0)[kpn−k+qn−k]=0$ is as follows:

$an(0)=−∑k=0n−1ak(0)[kpn−k+qn−k]n(n+1)$ with first couple of elements is as follows:

$a1(0)=−a0(0)[0+q1]1(1+1)=−11⋅2a2(0)=11⋅22⋅3a3(0)=−11⋅22⋅32⋅4$

And the values continues as such.

Hence, the first solution is $y1(x)=∑n=0∞(−1)nxnn!(n+1)!_$ .

According to the same theorem, since $r1−r2=1$ , the second solution is given by $y2(x)=ay1(x)ln|x|+|x|r2[1+∑n=1∞cn(r2)xn]$ .

$y2(x)=ay1(x)ln|x|+x−1[1+∑n=1∞cn(−1)xn]=ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn$

Where $a=limr→r2(r−r2)aN(r),N=r−r2$ since $N=1$ .

$aN(r)=a1(r)=−∑k=0N−1ak(r)[(r+k)pN−k+qN−k](r+N)(r+N+1)$

Substitute the limits in the above equation it becomes as follows:

$a1(r)=−a0(r)[(r+0)p1+q1](r+2)(r+1)=−[0+1](r+2)(r+1)=−1(r+2)(r+1)$

Thus, substitute the value of $a=limr→−1(r+1)−1(r+2)(r+1)=−1$ .

To calculate the value of $cn(0)$ substitute the value of $y2(x)=ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn$ into the initial differential equation.

First differentiate the equation $y2(x)=ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn$ .

$y2'(x)=ay1'(x)lnx+ay1(x)x−x−2+∑n=1∞ncn+1(−1)xn−1$

Again, differentiate the equation $y2'(x)=ay1'(x)lnx+ay1(x)x−x−2+∑n=1∞ncn+1(−1)xn−1$ .

$y2''(x)=ay1''(x)lnx+2ay1'(x)x−ay1(x)x+2x−3+∑n=1∞n(n−1)cn+1(−1)xn−2$

Now, the given differential equation becomes as follows:

$x2[ay1''(x)lnx+2ay1'(x)x−ay1(x)x+2x−3+∑n=1∞n(n−1)cn+1(−1)xn−2]+2x[ay1'(x)lnx+ay1(x)x−x−2+∑n=1∞ncn+1(−1)xn−1]+x[ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn]=0[2c2(−1)+c1(−1)−3a1(0)]x+∑n=1∞[−(2n+1)an(0)+n(n+1)]cn+1(−1)cn(−1)xn=0$

Equating the coefficient on both sides in the above equation.

${2c2(−1)+c1(−1)−3a1(0)=0−(2n+1)an(0)+n(n+1)cn+1(−1)cn(−1)=0$

Let $c2(−1)=1$ in the above equation this implies as follows:

$c2(−1)=3a1(0)−c1(−1)2=−H2+H12!c3(−1)=H3+H22⋅3$

$cn(−1)=(−1)n+1Hn+Hn−1n!(n−1)!$

Hence, the second solution is $y2(x)=−y1(x)lnx+x−1+[1+∑n=1∞(−1)n+1Hn+Hn−1n!(n−1)!xn]_$ .

Answer 6

Chapter 5.7, Problem 1P

To determine

To show: That the given differential equation has a regular singular point at $x=0$ is and determine two solutions for $x>0$ by using the given data.

Expert Solution & Answer

Answer to Problem 1P

The first solution of the differential equation is $y1(x)=∑n=0∞(−1)nxnn!(n+1)!_$ and second solution is $y2(x)=−y1(x)lnx+x−1[1+∑n=1∞(−1)n+1Hn+Hn−1n!(n−1)!xn]_$ .

Explanation of Solution

Theorem used:

Let $r1$ and $r2$ be the roots of the indicial equation $F(r)=r(r−1)+p0r+q0$ with $r1≥r2$ .

If $r1$ and $r2$ are real, then in either the interval $−ρ<x<0$ or the interval $0<x<ρ$ , there exists a solution of the form $y1(x)=|x|r1[1+∑n=1∞an(r1)xn]$ where $an(r1)$ are the given by recurrence relation with $a0=1$ and $r=r1$ .

If $r1−r2$ is not zero or a positive integer, then in either the interval $−ρ<x<0$ or the interval $0<x<ρ$ , there exists a second solution of the form $y2(x)=|x|r2[1+∑n=1∞an(r2)xn]$ where $an(r2)$ are also determined by the recurrence relation with $a0=1$ and $r=r2$ .

Further the power series converges at least for $|x|<ρ$ .

If $r1=r2$ , then the second solution is $y2(x)=y1(x)ln|x|+|x|r1[1+∑n=1∞bn(r1)xn]$ .

If $r1−r2=N$ , positive integer, then $y2(x)=ay1(x)ln|x|+|x|r2[1+∑n=1∞cn(r2)xn]$ .

Calculation:

The given differential equation is given as $x2y"+2xy'+xy=0$ .

Compare the equation $x2y"+2xy'+xy=0$ with $P(x)y"+Q(x)y'+R(x)y=0$ and note that $P(x)=x2$ , $Q(x)=2x$ and $R(x)=x$ .

Note that the singular points occur when $P(x0)=0$ .

Thus, the singular point of the given equation is $x0=0$ .

If $x0=0$ is a regular singular point, check whether the following limits are finite.

$p0=limx→0xQ(x)P(x)$ and $q0=limx→0x2R(x)P(x)$

For $x0=0$ , obtain the value of $p0$ as follows.

$p0=limx→0x2xx2=2$

Similarly, for $x0=0$ , obtain the value of $q0$ as follows.

$q0=limx→0x2x2x2=0$ .

Since both are finite, $x0=0$ is a regular singular point of the given differential equation.

Since $p0=2$ and $q0=0$ , the corresponding indicial equation is as follows:

$F(r)=r(r−1)+p0r+q0$

Substitute the value of $p0=2$ and $q0=0$ in $F(r)=r(r−1)+p0r+q0=0$ .

$F(r)=r(r−1)+2r+0=r2−r+2r=r2+r$

Let $F(r)=0$ in the above equation it becomes as follows:

$r2+r=0r(r+1)=0$

Therefore, the roots of the equation $r2−r+2r$ is $r=0,r=−1$ .

Hence, $x0$ is a regular singular point.

From the above Theorem, the first solution is given as $y1(x)=|x|r1[1+∑n=1∞an(r1)xn]$ .

Substitute the value of $r=0$ in $y1(x)=|x|r1[1+∑n=1∞an(r1)xn]$ .

$y1(x)=1+∑n=1∞an(0)xn$

Differentiate the equation $y1(x)=1+∑n=1∞an(0)xn$ .

$y1'(x)=∑n=1∞(n+1)an+1(0)xn$

The coefficients $an(0)$ , apart from $a0(0)=1$ are determined by the recurrence relation $F(r+n)an+∑k=1n−1ak[(r+k)pn−k+qn−k]=0$ in case of $r=r1$ .

$n(n+1)an(0)+∑k=1n−1ak(0)[kpn−k+qn−k]=0$ , where $∑n=0∞pnxn=xQ(x)P(x)=2$ and $∑n=0∞qnxn=x2R(x)P(x)=x$ .

Hence, the value of $p0=2,pn=0,∀n∈N$ and $q1=1,qn=0,∀n∈N0$ .

Factoring out $an(0)$ from the equation $n(n+1)an(0)+∑k=1n−1ak(0)[kpn−k+qn−k]=0$ is as follows:

$an(0)=−∑k=0n−1ak(0)[kpn−k+qn−k]n(n+1)$ with first couple of elements is as follows:

$a1(0)=−a0(0)[0+q1]1(1+1)=−11⋅2a2(0)=11⋅22⋅3a3(0)=−11⋅22⋅32⋅4$

And the values continues as such.

Hence, the first solution is $y1(x)=∑n=0∞(−1)nxnn!(n+1)!_$ .

According to the same theorem, since $r1−r2=1$ , the second solution is given by $y2(x)=ay1(x)ln|x|+|x|r2[1+∑n=1∞cn(r2)xn]$ .

$y2(x)=ay1(x)ln|x|+x−1[1+∑n=1∞cn(−1)xn]=ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn$

Where $a=limr→r2(r−r2)aN(r),N=r−r2$ since $N=1$ .

$aN(r)=a1(r)=−∑k=0N−1ak(r)[(r+k)pN−k+qN−k](r+N)(r+N+1)$

Substitute the limits in the above equation it becomes as follows:

$a1(r)=−a0(r)[(r+0)p1+q1](r+2)(r+1)=−[0+1](r+2)(r+1)=−1(r+2)(r+1)$

Thus, substitute the value of $a=limr→−1(r+1)−1(r+2)(r+1)=−1$ .

To calculate the value of $cn(0)$ substitute the value of $y2(x)=ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn$ into the initial differential equation.

First differentiate the equation $y2(x)=ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn$ .

$y2'(x)=ay1'(x)lnx+ay1(x)x−x−2+∑n=1∞ncn+1(−1)xn−1$

Again, differentiate the equation $y2'(x)=ay1'(x)lnx+ay1(x)x−x−2+∑n=1∞ncn+1(−1)xn−1$ .

$y2''(x)=ay1''(x)lnx+2ay1'(x)x−ay1(x)x+2x−3+∑n=1∞n(n−1)cn+1(−1)xn−2$

Now, the given differential equation becomes as follows:

$x2[ay1''(x)lnx+2ay1'(x)x−ay1(x)x+2x−3+∑n=1∞n(n−1)cn+1(−1)xn−2]+2x[ay1'(x)lnx+ay1(x)x−x−2+∑n=1∞ncn+1(−1)xn−1]+x[ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn]=0[2c2(−1)+c1(−1)−3a1(0)]x+∑n=1∞[−(2n+1)an(0)+n(n+1)]cn+1(−1)cn(−1)xn=0$

Equating the coefficient on both sides in the above equation.

${2c2(−1)+c1(−1)−3a1(0)=0−(2n+1)an(0)+n(n+1)cn+1(−1)cn(−1)=0$

Let $c2(−1)=1$ in the above equation this implies as follows:

$c2(−1)=3a1(0)−c1(−1)2=−H2+H12!c3(−1)=H3+H22⋅3$

$cn(−1)=(−1)n+1Hn+Hn−1n!(n−1)!$

Hence, the second solution is $y2(x)=−y1(x)lnx+x−1+[1+∑n=1∞(−1)n+1Hn+Hn−1n!(n−1)!xn]_$ .

Answer 7

Chapter 5.7, Problem 1P

To determine

To show: That the given differential equation has a regular singular point at $x=0$ is and determine two solutions for $x>0$ by using the given data.

Answer 8

Expert Solution & Answer

Answer to Problem 1P

The first solution of the differential equation is $y1(x)=∑n=0∞(−1)nxnn!(n+1)!_$ and second solution is $y2(x)=−y1(x)lnx+x−1[1+∑n=1∞(−1)n+1Hn+Hn−1n!(n−1)!xn]_$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Theorem used:

Let $r1$ and $r2$ be the roots of the indicial equation $F(r)=r(r−1)+p0r+q0$ with $r1≥r2$ .

If $r1$ and $r2$ are real, then in either the interval $−ρ<x<0$ or the interval $0<x<ρ$ , there exists a solution of the form $y1(x)=|x|r1[1+∑n=1∞an(r1)xn]$ where $an(r1)$ are the given by recurrence relation with $a0=1$ and $r=r1$ .

If $r1−r2$ is not zero or a positive integer, then in either the interval $−ρ<x<0$ or the interval $0<x<ρ$ , there exists a second solution of the form $y2(x)=|x|r2[1+∑n=1∞an(r2)xn]$ where $an(r2)$ are also determined by the recurrence relation with $a0=1$ and $r=r2$ .

Further the power series converges at least for $|x|<ρ$ .

If $r1=r2$ , then the second solution is $y2(x)=y1(x)ln|x|+|x|r1[1+∑n=1∞bn(r1)xn]$ .

If $r1−r2=N$ , positive integer, then $y2(x)=ay1(x)ln|x|+|x|r2[1+∑n=1∞cn(r2)xn]$ .

Calculation:

The given differential equation is given as $x2y"+2xy'+xy=0$ .

Compare the equation $x2y"+2xy'+xy=0$ with $P(x)y"+Q(x)y'+R(x)y=0$ and note that $P(x)=x2$ , $Q(x)=2x$ and $R(x)=x$ .

Note that the singular points occur when $P(x0)=0$ .

Thus, the singular point of the given equation is $x0=0$ .

If $x0=0$ is a regular singular point, check whether the following limits are finite.

$p0=limx→0xQ(x)P(x)$ and $q0=limx→0x2R(x)P(x)$

For $x0=0$ , obtain the value of $p0$ as follows.

$p0=limx→0x2xx2=2$

Similarly, for $x0=0$ , obtain the value of $q0$ as follows.

$q0=limx→0x2x2x2=0$ .

Since both are finite, $x0=0$ is a regular singular point of the given differential equation.

Since $p0=2$ and $q0=0$ , the corresponding indicial equation is as follows:

$F(r)=r(r−1)+p0r+q0$

Substitute the value of $p0=2$ and $q0=0$ in $F(r)=r(r−1)+p0r+q0=0$ .

$F(r)=r(r−1)+2r+0=r2−r+2r=r2+r$

Let $F(r)=0$ in the above equation it becomes as follows:

$r2+r=0r(r+1)=0$

Therefore, the roots of the equation $r2−r+2r$ is $r=0,r=−1$ .

Hence, $x0$ is a regular singular point.

From the above Theorem, the first solution is given as $y1(x)=|x|r1[1+∑n=1∞an(r1)xn]$ .

Substitute the value of $r=0$ in $y1(x)=|x|r1[1+∑n=1∞an(r1)xn]$ .

$y1(x)=1+∑n=1∞an(0)xn$

Differentiate the equation $y1(x)=1+∑n=1∞an(0)xn$ .

$y1'(x)=∑n=1∞(n+1)an+1(0)xn$

The coefficients $an(0)$ , apart from $a0(0)=1$ are determined by the recurrence relation $F(r+n)an+∑k=1n−1ak[(r+k)pn−k+qn−k]=0$ in case of $r=r1$ .

$n(n+1)an(0)+∑k=1n−1ak(0)[kpn−k+qn−k]=0$ , where $∑n=0∞pnxn=xQ(x)P(x)=2$ and $∑n=0∞qnxn=x2R(x)P(x)=x$ .

Hence, the value of $p0=2,pn=0,∀n∈N$ and $q1=1,qn=0,∀n∈N0$ .

Factoring out $an(0)$ from the equation $n(n+1)an(0)+∑k=1n−1ak(0)[kpn−k+qn−k]=0$ is as follows:

$an(0)=−∑k=0n−1ak(0)[kpn−k+qn−k]n(n+1)$ with first couple of elements is as follows:

$a1(0)=−a0(0)[0+q1]1(1+1)=−11⋅2a2(0)=11⋅22⋅3a3(0)=−11⋅22⋅32⋅4$

And the values continues as such.

Hence, the first solution is $y1(x)=∑n=0∞(−1)nxnn!(n+1)!_$ .

According to the same theorem, since $r1−r2=1$ , the second solution is given by $y2(x)=ay1(x)ln|x|+|x|r2[1+∑n=1∞cn(r2)xn]$ .

$y2(x)=ay1(x)ln|x|+x−1[1+∑n=1∞cn(−1)xn]=ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn$

Where $a=limr→r2(r−r2)aN(r),N=r−r2$ since $N=1$ .

$aN(r)=a1(r)=−∑k=0N−1ak(r)[(r+k)pN−k+qN−k](r+N)(r+N+1)$

Substitute the limits in the above equation it becomes as follows:

$a1(r)=−a0(r)[(r+0)p1+q1](r+2)(r+1)=−[0+1](r+2)(r+1)=−1(r+2)(r+1)$

Thus, substitute the value of $a=limr→−1(r+1)−1(r+2)(r+1)=−1$ .

To calculate the value of $cn(0)$ substitute the value of $y2(x)=ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn$ into the initial differential equation.

First differentiate the equation $y2(x)=ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn$ .

$y2'(x)=ay1'(x)lnx+ay1(x)x−x−2+∑n=1∞ncn+1(−1)xn−1$

Again, differentiate the equation $y2'(x)=ay1'(x)lnx+ay1(x)x−x−2+∑n=1∞ncn+1(−1)xn−1$ .

$y2''(x)=ay1''(x)lnx+2ay1'(x)x−ay1(x)x+2x−3+∑n=1∞n(n−1)cn+1(−1)xn−2$

Now, the given differential equation becomes as follows:

$x2[ay1''(x)lnx+2ay1'(x)x−ay1(x)x+2x−3+∑n=1∞n(n−1)cn+1(−1)xn−2]+2x[ay1'(x)lnx+ay1(x)x−x−2+∑n=1∞ncn+1(−1)xn−1]+x[ay1(x)lnx+x−1+∑n=1∞cn+1(−1)xn]=0[2c2(−1)+c1(−1)−3a1(0)]x+∑n=1∞[−(2n+1)an(0)+n(n+1)]cn+1(−1)cn(−1)xn=0$