Question

Question: In Problems , solve the equation.

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

Short answer

The solution of the given equation is $\mathbf{y}\mathbf{=}{\left(\mathbf{-}\mathbf{2}{\mathbf{x}}^3\mathbf{+}{\mathbf{Cx}}^2\right)}^{-1}$.

Step-by-step solution

Given information and simplification

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

Multiply by y^-2 in equation (1),

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

Now substitute u = y^-1 differentiate with respect to x.

$\begin{array}{c}\frac{\mathrm{du}}{\mathrm{dx}}\mathbf{=}\mathbf{-}{\mathbf{y}}^{-2}\frac{\mathrm{dy}}{\mathrm{dx}}\\ \mathbf{-}\frac{\mathrm{du}}{\mathrm{dx}}\mathbf{=}{\mathbf{y}}^{-2}\frac{\mathrm{dy}}{\mathrm{dx}}\end{array}$

Substitute in equation (2).

$\begin{array}{c}\mathbf{-}\frac{\mathrm{du}}{\mathrm{dx}}\mathbf{+}\frac{2}{x}\mathbf{u}\mathbf{=}\mathbf{2}{\mathbf{x}}^2\\ \frac{\mathrm{du}}{\mathrm{dx}}\mathbf{-}\frac{2}{x}\mathbf{u}\mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{x}}^2\cdot \cdot \cdot \cdot \cdot \cdot \left(3\right)\end{array}$

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

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

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

Evaluation method

Multiply x^-2 in equation (3) on both sides.

$\begin{array}{c}{\mathbf{x}}^{-2}\frac{\mathrm{du}}{\mathrm{dx}}\mathbf{-}{\mathbf{x}}^{-2}\frac{2}{x}\mathbf{u}\mathbf{=}\mathbf{-}\mathbf{2}\\ \frac{1}{{\mathbf{x}}^2}\frac{\mathrm{du}}{\mathrm{dx}}\mathbf{-}\frac{2}{{\mathbf{x}}^3}\mathbf{u}\mathbf{=}\mathbf{-}\mathbf{2}\\ \frac{d}{\mathrm{dx}}\left[\frac{1}{{\mathbf{x}}^2}\mathbf{u}\right]\mathbf{=}\mathbf{-}\mathbf{2}\end{array}$

Now integrate the equation on both sides.

$\begin{array}{c}\int \frac{d}{\mathrm{dx}}\left[\frac{1}{{\mathbf{x}}^2}\mathbf{u}\right]\mathbf{dx}\mathbf{=}\int \mathbf{-}\mathbf{2}\mathbf{dx}\\ \frac{1}{{\mathbf{x}}^2}\mathbf{u}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{x}\mathbf{+}\mathbf{C}\\ \mathbf{u}\mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{x}}^3\mathbf{+}{\mathbf{Cx}}^2\end{array}$

Substitute $\mathbf{u}\mathbf{=}{\mathbf{y}}^{-1}$, we get,

$\begin{array}{c}{\mathbf{y}}^{-1}\mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{x}}^3\mathbf{+}{\mathbf{Cx}}^2\\ \mathbf{y}\mathbf{=}{\left(\mathbf{-}\mathbf{2}{\mathbf{x}}^3\mathbf{+}{\mathbf{Cx}}^2\right)}^{-1}\end{array}$

So, the solution is $\mathbf{y}\mathbf{=}{\left(\mathbf{-}\mathbf{2}{\mathbf{x}}^3\mathbf{+}{\mathbf{Cx}}^2\right)}^{-1}$