Chapter D2, Problem 1RE

a.

To determine

To explain: Whether the statement “The equation $y^{″}+2y^{\prime }-ty=0$ is second order, linear, and homogeneous” is true or not.

Answer to Problem 1RE

The statement is true.

Explanation of Solution

Given:

The differential equation is $y^{″}+2y^{\prime }-ty=0$.

The highest derivative occur in the given equation is two.

Therefore the equation is second order.

Here, it is observed that the differential equation does not contain any product of the function and its derivative and all the power of the derivative are one.

Thus, the nonhomogeneous part of the equation $f\left(t\right)=0$.

Therefore, the differential equation is linear and nonhomogeneous.

Thus, the given statement is true.

b.

To determine

To explain: Whether the statement “The equation $y^{″}+2y^{\prime }-y^2=3$ is second order, linear, homogeneous” is true or not.

Answer to Problem 1RE

The given statement is false.

Explanation of Solution

Given:

The differential equation is $y^{″}+2y^{\prime }-y^2=3$.

The highest derivative occur in the given equation is two. Therefore the equation is second order.

Here, it is observed that the differential equation contains the term $y^2$.

Therefore the equation is nonlinear.

Thus, the nonhomogeneous part of the equation $f\left(t\right)=3$.

That is, the differential equation is nonhomogeneous.

Hence, the given statement is false.

c.

To determine

To explain: Whether the statement “To solve the equation $t^2{y}^{″}+2t{y}^{\prime }+4y=0$, use the trial solution $y=e^{rt}$” is true or not; If true find r.

Answer to Problem 1RE

The statement is false.

Explanation of Solution

Consider the given differential equation $t^2{y}^{″}+2t{y}^{\prime }+4y=0$.

The above equation of the form $t^2{y}^{″}+a{y}^{\prime }+by=0$.

Therefore, it is a Cauchy-Euler equation.

Thus, the trial solution is of the form $t^p$.

That is, cannot solve the equation by using trial solution $e^{rt}$.

Hence, the given statement is false.

d.

To determine

To explain: Whether the statement “To find the particular solution of the given differential equation, using the trial solution as $y_p=A{e}^t$” is true or not

Answer to Problem 1RE

The statement is false.

Explanation of Solution

Consider the given differential equation $y^{″}-2y^{\prime }+y=e^t$.

Here, the homogeneous part of the equation is $y^{″}-2y^{\prime }+y=0$.

The characteristic polynomial of the homogeneous part of the equation is, $r^2-2r+1=0$

$\left(r+1\right)\left(r+1\right)=0$.

Now, the general solution of the homogeneous part of the equation is $C_1{e}^t+C_{2}t{e}^t$.

Therefore, the function $A{e}^t,Bt{e}^t$ is also a solution of the homogeneous part of the equation.

Thus, the trial solution is $A{t}^2{e}^t$ but the function $y=A{e}^t$ is not a trial solution for the equation.

Hence, the statement is false.

e.

To determine

To explain: Whether the statement “The function $y=0$ is always a solution of second order linear homogeneous equation” is true or not.

Answer to Problem 1RE

The statement is true.

Explanation of Solution

Consider the general form of the second order linear homogeneous equation, $y^{″}+p\left(t\right)y^{\prime }+q\left(t\right)y=0$.

In the above equation, there is no separate term of function of t and each additive terms associate with $y$ or derivative of y.

Therefore, each additive term will be zero.

Thus, $y=0$ is a solution of second order homogeneous equation.

Hence, the given statement is true.