In Problems , solve the initial value problem.

$\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{-}\mathbf{4}\mathbf{y}\mathbf{=}\mathbf{2}{\mathbf{xy}}^{\mathbf{2}}\mathbf{,} \mathbf{y}\left(\mathbf{0}\right)\mathbf{=}\mathbf{-}\mathbf{4}$

Given that, $\frac{\mathrm{dy}}{\mathrm{dx}}-4\mathrm{y}=2{\mathrm{xy}}^2, \mathrm{y}\left(0\right)=-4······\left(1\right)$

Evaluate the equation (1).

$\frac{\mathrm{dy}}{\mathrm{dx}}-4\mathrm{y}=2{\mathrm{xy}}^2$

Multiply $\frac{1}{{\mathrm{y}}^2}$on both sides.

$\frac{1}{{\mathrm{y}}^2}\frac{\mathrm{dy}}{\mathrm{dx}}-4\frac{1}{\mathrm{y}}=2\mathrm{x}······\left(2\right)$

Let us assume $\mathrm{v}=\frac{1}{\mathrm{y}}$. Differentiate with respect to x.

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

Substitute the values in equation (2)

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

Let P = 4.

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

$$\begin{gathered} \mu \left(\mathrm{x}\right)={\mathrm{e}}^{\int \mathrm{P}\left(\mathrm{x}\right)\mathrm{dx}} \\ ={\mathrm{e}}^{\int 4\mathrm{dx}} \\ ={\mathrm{e}}^{4\mathrm{x}} \end{gathered}$$

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

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

Now integrate the equation on both sides.

$$\begin{gathered} \int \frac{\mathrm{d}}{\mathrm{dx}}\left[{\mathrm{e}}^{4\mathrm{x}}\mathrm{v}\right]\mathrm{dx}=-\int 2\mathrm{x}{\mathrm{e}}^{4\mathrm{x}}\mathrm{dx} \\ {\mathrm{e}}^{4\mathrm{x}}\mathrm{v}=\frac{1}{8}{\mathrm{e}}^{4\mathrm{x}}-\frac{1}{2}\mathrm{x}{\mathrm{e}}^{4\mathrm{x}}+{\mathrm{C}}_1 \\ \mathrm{v}=\frac{1}{8}-\frac{1}{2}\mathrm{x}+{\mathrm{Ce}}^{-4\mathrm{x}} \end{gathered}$$

Substitute the value of v.

$$\begin{gathered} \frac{1}{\mathrm{y}}=\frac{1}{8}-\frac{1}{2}\mathrm{x}+{\mathrm{Ce}}^{-4\mathrm{x}} \\ \mathrm{y}=\frac{1}{\frac{1}{8}-\frac{1}{2}\mathrm{x}+{\mathrm{Ce}}^{-4\mathrm{x}}} \\ \mathrm{y}=\frac{8}{1-4\mathrm{x}+8{\mathrm{Ce}}^{-4\mathrm{x}}}······\left(3\right) \end{gathered}$$

So, the solution is found.

Given that, $\mathrm{y}\left(0\right)=-4$.

Then, x = 0 and y = -4.

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

role="math" localid="1664254188356" $$\begin{gathered} \mathrm{y}=\frac{8}{1-4\mathrm{x}+8{\mathrm{Ce}}^{-4\mathrm{x}}} \\ -4=\frac{8}{1-4\left(0\right)+8{\mathrm{Ce}}^{-4\left(0\right)}} \\ -4=\frac{8}{1+8\mathrm{C}} \\ -4-32\mathrm{C}=8 \\ \mathrm{C}=-\frac{12}{32} \\ =-\frac{3}{8} \end{gathered}$$

Substitute the value of C in equation (3).

$$\begin{gathered} \mathrm{y}=\frac{8}{1-4\mathrm{x}+8\left(-\frac{3}{8}\right){\mathrm{e}}^{-4\mathrm{x}}} \\ \mathrm{y}=\frac{8}{1-4\mathrm{x}-3{\mathrm{e}}^{-4\mathrm{x}}} \end{gathered}$$

So, the solution is role="math" localid="1664254272621" $\mathbf{y}\mathbf{=}\frac{\mathbf{8}}{\mathbf{1}\mathbf{-}\mathbf{4}\mathbf{x}\mathbf{-}\mathbf{3}{\mathbf{e}}^{\mathbf{-}\mathbf{4}\mathbf{x}}}$