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

$\left(\mathbf{2}{\mathbf{y}}^2\mathbf{+}\mathbf{4}{\mathbf{x}}^2\right)\mathbf{dx}\mathbf{-}\mathbf{xydy}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\left(1\right)\mathbf{=}\mathbf{-}\mathbf{2}$

Given that, $\left(\mathbf{2}{\mathbf{y}}^2\mathbf{+}\mathbf{4}{\mathbf{x}}^2\right)\mathbf{dx}\mathbf{-}\mathbf{xy}\mathbf{dy}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{y}\left(1\right)\mathbf{=}\mathbf{-}\mathbf{2}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

Let us check whether the given equation is exact or not.

Then, $\mathbf{M}\mathbf{=}\mathbf{2}{\mathbf{y}}^2\mathbf{+}\mathbf{4}{\mathbf{x}}^2\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{-}\mathbf{xy}$.

Differentiate the value of M and N.

$\begin{array}{l}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}4\mathbf{y}\\ \frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{=}\mathbf{-}\mathbf{y}\end{array}$

So, the given equation is not exact. Then, find the special integrating factor,

$\begin{array}{c}\frac{\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{-}\frac{\partial \mathbf{N}}{\partial \mathbf{x}}}{N}\mathbf{=}\frac{\mathbf{-}4\mathbf{y}\mathbf{-}\mathbf{y}}{\mathrm{xy}}\\ \mathbf{=}\frac{\mathbf{y}\left(-5\right)}{\mathrm{xy}}\\ \mathbf{=}\mathbf{-}\frac{5}{x}\end{array}$

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

$\begin{array}{c}\mu \left(x\right)\mathbf{=}{\mathbf{e}}^{\mathbf{-}\int \frac{5}{x}\mathbf{dx}}\\ \mathbf{=}{\mathbf{e}}^{\mathbf{-}5\mathbf{ln}\left|x\right|}\\ \mathbf{=}{\mathbf{x}}^{\mathbf{-}5}\end{array}$

Multiply ${\mathbf{x}}^{\mathbf{-}5}$in equation (1) on both sides.

$\begin{array}{c}{\mathbf{x}}^{\mathbf{-}5}\left(\mathbf{2}{\mathbf{y}}^2\mathbf{+}\mathbf{4}{\mathbf{x}}^2\right)\mathbf{dx}\mathbf{-}{\mathbf{xx}}^{\mathbf{-}5}\mathbf{y}\mathbf{dy}\mathbf{=}\mathbf{0}\\ \left(\mathbf{2}{\mathbf{x}}^{-5}{\mathbf{y}}^2\mathbf{+}\mathbf{4}{\mathbf{x}}^{-3}\right)\mathbf{dx}\mathbf{-}{\mathbf{x}}^{-4}\mathbf{y}\mathbf{dy}\mathbf{=}\mathbf{0}\end{array}$

Now again check whether the founded equation is exact or not.

$\mathbf{M}\mathbf{=}\mathbf{2}{\mathbf{x}}^{-5}{\mathbf{y}}^2\mathbf{+}\mathbf{4}{\mathbf{x}}^{-3}\mathbf{,}\mathbf{N}\mathbf{=}{\mathbf{x}}^{-4}\mathbf{y}$

Differentiate the value of M and N.

$\begin{array}{l}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\mathbf{4}{\mathbf{x}}^{-5}\mathbf{y}\\ \frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{=}\mathbf{4}{\mathbf{x}}^{-5}\mathbf{y}\end{array}$

Therefore, the founded equation is exact.

Now, let us assume $\mathbf{M}\mathbf{=}\frac{\partial \mathbf{F}}{\partial \mathbf{x}}\mathbf{=}\mathbf{2}{\mathbf{x}}^{-5}{\mathbf{y}}^2\mathbf{+}\mathbf{4}{\mathbf{x}}^{-3}$.

Integrate on both sides.

$\begin{array}{c}\mathbf{F}\mathbf{=}\int \left(\mathbf{2}{\mathbf{x}}^{-5}{\mathbf{y}}^2\mathbf{+}\mathbf{4}{\mathbf{x}}^{-3}\right)\mathbf{dx}\\ \mathbf{=}\mathbf{-}\mathbf{2}{\mathbf{x}}^{-2}\mathbf{-}\frac{{\mathbf{y}}^2{\mathbf{x}}^{-4}}{2}\mathbf{+}\mathbf{g}\left(y\right)\end{array}$

Differentiate the F with respect to y.

$\begin{array}{c}\frac{\partial \mathbf{F}}{\partial \mathbf{y}}\mathbf{=}\mathbf{-}{\mathbf{yx}}^{-4}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(y\right)\\ =N\end{array}$

Equalize the values of N.

$\begin{array}{c}\mathbf{-}{\mathbf{yx}}^{-4}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}\mathbf{-}{\mathbf{yx}}^{-4}\\ \mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}0\end{array}$

Integrate on both sides.

$\begin{array}{c}\int \mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}\int 0\mathbf{dy}\\ \mathbf{g}\left(y\right)\mathbf{=}{\mathbf{C}}_1\end{array}$

Substitute in the equation of F.

$\begin{array}{c}\mathbf{-}\mathbf{2}{\mathbf{x}}^{-2}\mathbf{-}\frac{{\mathbf{y}}^2{\mathbf{x}}^{-4}}{2}\mathbf{+}{\mathbf{C}}_1\mathbf{=}\mathbf{0}\\ \mathbf{-}\mathbf{2}{\mathbf{x}}^{-2}\mathbf{-}\frac{{\mathbf{y}}^2{\mathbf{x}}^{-4}}{2}\mathbf{=}\mathbf{C}\cdot \cdot \cdot \cdot \cdot \cdot \left(2\right)\end{array}$

So, the solution is found.

Given that, $\mathbf{y}\left(1\right)\mathbf{=}\mathbf{-}\mathbf{2}$.

Then, x = 1 and y = -2.

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

Substitute the value of C in equation (2).

$\begin{array}{c}\mathbf{-}\mathbf{2}{\mathbf{x}}^{-2}\mathbf{-}\frac{{\mathbf{y}}^2{\mathbf{x}}^{-4}}{2}\mathbf{=}\mathbf{-}\mathbf{4}\\ \mathbf{-}\mathbf{4}{\mathbf{x}}^{-2}\mathbf{-}{\mathbf{y}}^2{\mathbf{x}}^{-4}\mathbf{=}\mathbf{-}\mathbf{8}\\ \mathbf{4}{\mathbf{x}}^2\mathbf{+}{\mathbf{y}}^2\mathbf{=}\mathbf{8}{\mathbf{x}}^4\\ {\mathbf{y}}^2\mathbf{=}\mathbf{8}{\mathbf{x}}^4\mathbf{-}\mathbf{4}{\mathbf{x}}^2\\ \mathbf{y}\mathbf{=}\mathbf{4}{\mathbf{x}}^2\left(\mathbf{2}{\mathbf{x}}^2\mathbf{-}\mathbf{1}\right)\\ \mathbf{y}\mathbf{=}\mathbf{\pm }\sqrt{\mathbf{4}{\mathbf{x}}^2\left(\mathbf{2}{\mathbf{x}}^2\mathbf{-}\mathbf{1}\right)}\\ \mathbf{y}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{x}\sqrt{\mathbf{2}{\mathbf{x}}^2\mathbf{-}\mathbf{1}}\end{array}$

Since the initial value of y is negative. So that we get the negative value.

So, the solution is $\mathbf{y}\mathbf{=}\mathbf{-}\mathbf{2}\mathbf{x}\sqrt{\mathbf{2}{\mathbf{x}}^2\mathbf{-}\mathbf{1}}$