91Ó°ÊÓ

Given that, $\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{2\mathrm{y}}{\mathrm{x}}={\mathrm{x}}^{-1}{\mathrm{y}}^{-1},    \mathrm{y}\left(1\right)=3······\left(1\right)$

Evaluate the equation (1).

$\frac{\mathrm{dy}}{\mathrm{dx}}-\frac{2\mathrm{y}}{\mathrm{x}}={\mathrm{x}}^{-1}{\mathrm{y}}^{-1}$

Multiply y on both sides.

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

Let us assume $\mathrm{v}={\mathrm{y}}^2$. Differentiate with respect to x.

$\frac{\mathrm{dv}}{\mathrm{dx}}=2\mathrm{y}\frac{\mathrm{dy}}{\mathrm{dx}}$

Substitute the values in equation (2)

$$\begin{gathered} \frac{1}{2}\frac{\mathrm{dv}}{\mathrm{dx}}-\frac{2\mathrm{v}}{\mathrm{x}}={\mathrm{x}}^{-1} \\ \frac{\mathrm{dv}}{\mathrm{dx}}-\frac{4\mathrm{v}}{\mathrm{x}}=2{\mathrm{x}}^{-1} \end{gathered}$$

Let $\mathrm{P}\left(\mathrm{x}\right)=-\frac{4}{\mathrm{x}}$.

Then find the value of$\mathrm{μ}\left(\mathbf{x}\right)$.

$$\begin{gathered} \mathrm{μ}\left(\mathrm{x}\right)={\mathrm{e}}^{∫\mathrm{P}\left(\mathrm{x}\right)\mathrm{dx}} \\ ={\mathrm{e}}^{∫-\frac{4}{\mathrm{x}}\mathrm{dx}} \\ ={\mathrm{e}}^{-4\ln \left|\mathrm{x}\right|} \\ ={\mathrm{x}}^{-4} \end{gathered}$$

Multiply $\frac{1}{{\mathrm{x}}^4}$on both sides.

$$\begin{gathered} \frac{1}{{\mathrm{x}}^4}\frac{\mathrm{dv}}{\mathrm{dx}}-\frac{4\mathrm{v}}{{\mathrm{x}}^5}=\frac{2}{{\mathrm{x}}^5} \\ \frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{1}{{\mathrm{x}}^4}\mathrm{v}\right]=\frac{2}{{\mathrm{x}}^5} \end{gathered}$$

Now integrate the equation on both sides.

$$\begin{gathered} ∫\frac{\mathrm{d}}{\mathrm{dx}}\left[\frac{1}{{\mathrm{x}}^4}\mathrm{v}\right]\mathrm{dx}=∫\frac{2}{{\mathrm{x}}^5}\mathrm{dx} \\ \frac{1}{{\mathrm{x}}^4}\mathrm{v}=-\frac{1}{2{\mathrm{x}}^4}+{\mathrm{C}}_1 \\ \mathrm{v}=-\frac{1}{2}+{\mathrm{Cx}}^4 \end{gathered}$$

Substitute the value of v.

$$\begin{gathered} {\mathrm{y}}^2=-\frac{1}{2}+{\mathrm{Cx}}^4 \\ \mathrm{y}=\sqrt{-\frac{1}{2}+{\mathrm{Cx}}^4}······\left(3\right) \end{gathered}$$

So, the solution is found.

Given that, $\mathrm{y}\left(1\right)=3$.

Then, x = 1 and y = 3.

Substitute the value in equation (3) to get the value of C.

$$\begin{gathered} \mathrm{y}=\sqrt{-\frac{1}{2}+{\mathrm{Cx}}^4} \\ 3=\sqrt{-\frac{1}{2}+\mathrm{C}} \\ \mathrm{C}=\frac{19}{2} \end{gathered}$$

$$\begin{gathered} \mathrm{y}=\sqrt{-\frac{1}{2}+\frac{19}{2}{\mathrm{x}}^4} \\ \mathrm{y}=\sqrt{\frac{19{\mathrm{x}}^4-1}{2}} \end{gathered}$$

So, the solution isrole="math" localid="1664253082555" $\mathbf{y}\mathbf{=}\sqrt{\frac{\mathbf{19}{\mathbf{x}}^{\mathbf{4}}\mathbf{-}\mathbf{1}}{\mathbf{2}}}$