Question

Question: In Problems 31-40, solve the initial value problem.

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{-}\frac{2y}{x}\mathbf{=}{\mathbf{x}}^2\mathbf{cosx}\mathbf{,}\mathbf{y}\left(\pi \right)\mathbf{=}\mathbf{2}$

Short answer

The solution of the given equation is $\mathbf{y}\mathbf{=}{\mathbf{x}}^2\mathbf{sin}\mathbf{x}\mathbf{+}\frac{\mathbf{2}{\mathbf{x}}^2}{{\pi }^2}$.

Step-by-step solution

Given information and simplification

Given that, $\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{-}\frac{2y}{x}\mathbf{=}{\mathbf{x}}^2\mathbf{cos}\mathbf{x}\mathbf{,}\mathbf{y}\left(\pi \right)\mathbf{=}\mathbf{2}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Let $\mathbf{P}\left(x\right)\mathbf{=}\mathbf{-}\frac{2}{x}$.

Find the value of$\mu \left(x\right)$.

$\begin{array}{c}\mu \left(x\right)\mathbf{=}{\mathbf{e}}^{\int \mathbf{P}\left(x\right)\mathbf{dx}}\\ \mathbf{=}{\mathbf{e}}^{\mathbf{-}\int \frac{2}{x}\mathbf{dx}}\\ \mathbf{=}{\mathbf{e}}^{\mathbf{-}\mathbf{2}\mathbf{ln}\left|x\right|}\\ \mathbf{=}\frac{1}{{\mathbf{x}}^2}\end{array}$

Multiply$\frac{1}{{\mathbf{x}}^2}$ in equation (1) on both sides.

$\begin{array}{c}\frac{1}{{\mathbf{x}}^2}\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{-}\frac{2y}{{\mathbf{x}}^3}\mathbf{=}\mathbf{cos}\mathbf{x}\\ \frac{\mathrm{dy}}{\mathrm{dx}}\left[\frac{1}{{\mathbf{x}}^2}\mathbf{y}\right]\mathbf{=}\mathbf{cos}\mathbf{x}\end{array}$

integration method

Now integrate the equation on both sides.

${\begin{array}{c}\int \frac{\mathrm{dy}}{\mathrm{dx}}\left[\frac{1}{{\mathbf{x}}^2}\mathbf{y}\right]\mathbf{dx}\mathbf{=}\int \mathbf{cos}\mathbf{x}\mathbf{dx}\\ \frac{1}{{\mathbf{x}}^2}\mathbf{y}\mathbf{=}\mathbf{sin}\mathbf{x}\mathbf{+}{\mathbf{C}}_1\\ \mathbf{y}\mathbf{=}{\mathbf{x}}^2\mathbf{sin}\mathbf{x}\mathbf{+}{\mathbf{x}}^2\mathbf{C}\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)\end{array}}_{}$

So, the solution is found.

Find the initial value

Given that, $\mathbf{y}\left(\pi \right)\mathbf{=}\mathbf{2}$.

Then,$\mathbf{x}\mathbf{=}\pi$ and y = 2.

Substitute the value in equation (2) to get the value of C.

$\begin{array}{c}\mathbf{y}\mathbf{=}{\mathbf{x}}^2\mathbf{sin}\mathbf{x}\mathbf{+}{\mathbf{x}}^2\mathbf{C}\\ \mathbf{2}\mathbf{=}{\pi }^2\mathbf{sin}\left(\pi \right)\mathbf{+}{\pi }^2\mathbf{C}\\ \frac{2}{{\pi }^2}\mathbf{=}\mathbf{C}\end{array}$

Substitute the value of C in equation (2).

$\mathbf{y}\mathbf{=}{\mathbf{x}}^2\mathbf{sin}\mathbf{x}\mathbf{+}\frac{\mathbf{2}{\mathbf{x}}^2}{{\pi }^2}$

So, the solution is $\mathbf{y}\mathbf{=}{\mathbf{x}}^2\mathbf{sin}\mathbf{x}\mathbf{+}\frac{\mathbf{2}{\mathbf{x}}^2}{{\pi }^2}$