Chapter 6.1, Problem 78E

(a)

To determine

To find the solution of second order non homogenous initial value problem.

Answer to Problem 78E

  $y=2x^4-5x^2+5x+3$

Explanation of Solution

Given: Second order non homogenous differential equation is

  $\frac{d^{2}y}{d{x}^2}=24x^2-10$ and initial values are

  $x=1,\frac{dy}{dx}=3$ and $y=5$ .

Calculation:

  $\begin{array}{l}\frac{d^{2}y}{d{x}^2}=24x^2-10\\ \Rightarrow \frac{dy}{dx}={\displaystyle \int \left(24x^2-10\right)}\\ ={\displaystyle \int 24x^2}-{\displaystyle \int 10}\\ =24\times \frac{\left(x^3\right)}{3}-10x+c_1\\ \Rightarrow \frac{dy}{dx}=8x^3-10x+c_1\mathrm{..........}\left(1\right)\end{array}$

Where $c_1$ is integration constant.

  $\because x=1,\frac{dy}{dx}=3$

  $\begin{array}{l}\therefore 3=\left(8\times 1\right)-\left(10\times 1\right)+c_1\\ \Rightarrow 3=8-10+c_1\\ \Rightarrow c_1=5\end{array}$

Substitute $c_1$ value in equation $\left(1\right)$ then

  $\frac{dy}{dx}=8x^3-10x+5\mathrm{.......}\left(2\right)$

Now

  $\begin{array}{l}y={\displaystyle \int \left(8x^3-10x+5\right)}\\ ={\displaystyle \int 8x^3}-{\displaystyle \int 10x}+{\displaystyle \int 5}\\ =8\times \frac{x^4}{4}-10\times \frac{x^2}{2}+5x\\ =2x^4-5x^2+5x+c_2\mathrm{........}\left(3\right)\end{array}$

  $\because x=1,y=5$

  $\begin{array}{l}5=2\times \left(1\right)-5\times \left(1\right)+5\times \left(1\right)+c_2\\ \Rightarrow c_2=3\end{array}$

Substitute $c_2$ value in equation $\left(3\right)$ then

  $y=2x^4-5x^2+5x+3$

Conclusion:

  $y=2x^4-5x^2+5x+3$

(b)

To determine

To find the solution of second order non homogenous initial value problem.

Answer to Problem 78E

  $y=\sin x-\cos x+x+1$

Explanation of Solution

Given: Second order non homogenous differential equation is

  $\frac{d^{2}y}{d{x}^2}=\cos x-\sin x$ and initial values are

  $x=0,\frac{dy}{dx}=2,y=0$

Calculation:

  $\begin{array}{l}\frac{d^{2}y}{d{x}^2}=\cos x-\sin x\\ \Rightarrow \frac{dy}{dx}={\displaystyle \int \left(\cos x-\sin x\right)}\\ ={\displaystyle \int \cos x}-{\displaystyle \int \sin x}\\ =\sin x+\cos x+c_1\mathrm{.......}\left(1\right)\end{array}$

  $\because x=0,\frac{dy}{dx}=2$

  $\begin{array}{l}\therefore 2=\sin \left(0\right)+\cos \left(0\right)+c_1\\ \Rightarrow 2=0+1+c_1\\ \Rightarrow c_1=1\end{array}$

Substitute $c_1$ value in equation $\left(1\right)$ then

  $\frac{dy}{dx}=\sin x+\cos x+1\mathrm{...........}\left(2\right)$

Now

  $\begin{array}{l}y={\displaystyle \int \left(\sin x+\cos x+1\right)}\\ ={\displaystyle \int \sin x}+{\displaystyle \int \cos x}+{\displaystyle \int 1}\\ =-\cos x+\sin x+x+c_2\mathrm{.......}\left(3\right)\end{array}$

  $\because x=0,y=0$ then

  $\begin{array}{l}0=-\cos \left(0\right)+\sin \left(0\right)+0+c_2\\ \Rightarrow 0=-1+0+c_2\\ \Rightarrow c_2=1\end{array}$

Substitute $c_2$ value in equation $\left(3\right)$ then

  $y=\sin x-\cos x+x+1$

Conclusion:

  $y=\sin x-\cos x+x+1$