Question

Question: In Problems 1 - 30, solve the equation.

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{-}\frac{y}{x}\mathbf{=}{\mathbf{x}}^2\mathbf{sin}\mathbf{2}\mathbf{x}$

Short answer

The solution of the given equation is $\mathbf{y}\mathbf{=}\mathbf{-}\frac{{\mathbf{x}}^2\mathbf{cos}\mathbf{2}\mathbf{x}}{2}\mathbf{+}\frac{\mathbf{x}\mathbf{sin}\mathbf{2}\mathbf{x}}{4}\mathbf{+}\mathbf{Cx}$.

Step-by-step solution

Given information and simplification

Given that,

$\frac{\mathrm{dy}}{\mathrm{dx}}\mathbf{-}\frac{y}{x}\mathbf{=}{\mathbf{x}}^2\mathbf{sin}\mathbf{2}\mathbf{x}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Let $\mathbf{P}\left(x\right)\mathbf{=}\mathbf{-}\frac{1}{x}$.

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

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

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

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

Now integrate the equation on both sides.

$\int \frac{d}{\mathrm{dx}}\left[\frac{1}{x}\mathbf{y}\right]\mathbf{dx}\mathbf{=}\int \mathbf{x}\mathbf{sin}\mathbf{2}\mathbf{x}\mathbf{dx}\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)$

Evaluation method

Find the value of $\int \mathbf{x}\mathbf{sin}\mathbf{2}\mathbf{x}\mathbf{dx}$ separately.

Let us take $\mathbf{u}\mathbf{=}\mathbf{x}\mathbf{,}\mathbf{dv}\mathbf{=}\mathbf{sin}\mathbf{2}\mathbf{x}\mathbf{dx}$.

$\mathbf{du}\mathbf{=}\mathbf{dx}\mathbf{,}\mathbf{v}\mathbf{=}\mathbf{-}\frac{\mathbf{cos}\mathbf{2}\mathbf{x}}{2}$

Use the integration by parts formula.

$\begin{array}{c}\int \mathbf{x}\mathbf{sin}\mathbf{2}\mathbf{x}\mathbf{dx}\mathbf{=}\mathbf{-}\frac{\mathbf{x}\mathbf{cos}\mathbf{2}\mathbf{x}}{2}\mathbf{+}\frac{1}{2}\int \mathbf{cos}\mathbf{2}\mathbf{x}\mathbf{dx}\\ \mathbf{=}\mathbf{-}\frac{\mathbf{x}\mathbf{cos}\mathbf{2}\mathbf{x}}{2}\mathbf{+}\frac{\mathbf{sin}\mathbf{2}\mathbf{x}}{4}\mathbf{+}{\mathbf{C}}_1\end{array}$

Now substitute in equation (2)

$\begin{array}{c}\int \frac{d}{\mathrm{dx}}\left[\frac{1}{x}\mathbf{y}\right]\mathbf{dx}\mathbf{=}\int \mathbf{x}\mathbf{sin}\mathbf{2}\mathbf{x}\mathbf{dx}\\ \frac{1}{x}\mathbf{y}\mathbf{=}\mathbf{-}\frac{\mathbf{x}\mathbf{cos}\mathbf{2}\mathbf{x}}{2}\mathbf{+}\frac{\mathbf{sin}\mathbf{2}\mathbf{x}}{4}\mathbf{+}{\mathbf{C}}_1\\ \mathbf{y}\mathbf{=}\mathbf{-}\frac{{\mathbf{x}}^2\mathbf{cos}\mathbf{2}\mathbf{x}}{2}\mathbf{+}\frac{\mathbf{x}\mathbf{sin}\mathbf{2}\mathbf{x}}{4}\mathbf{+}\mathbf{Cx}\end{array}$

So, the solution is $\mathbf{y}\mathbf{=}\mathbf{-}\frac{{\mathbf{x}}^2\mathbf{cos}\mathbf{2}\mathbf{x}}{2}\mathbf{+}\frac{\mathbf{x}\mathbf{sin}\mathbf{2}\mathbf{x}}{4}\mathbf{+}\mathbf{Cx}$