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

$\left({\mathbf{y}}^3\mathbf{+}\mathbf{4}{\mathbf{e}}^x\mathbf{y}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{2}{\mathbf{e}}^x\mathbf{+}\mathbf{3}{\mathbf{y}}^2\right)\mathbf{dy}\mathbf{=}\mathbf{0}$

Given that, $\left({\mathbf{y}}^3\mathbf{+}\mathbf{4}{\mathbf{e}}^x\mathbf{y}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{2}{\mathbf{e}}^x\mathbf{+}\mathbf{3}{\mathbf{y}}^2\right)\mathbf{dy}\mathbf{=}\mathbf{0}\cdot \cdot \cdot \cdot \cdot \cdot \left(1\right)$

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

Then, $\mathbf{M}\mathbf{=}{\mathbf{y}}^3\mathbf{+}\mathbf{4}{\mathbf{e}}^x\mathbf{y}\mathbf{,}\mathbf{N}\mathbf{=}\mathbf{2}{\mathbf{e}}^x\mathbf{+}\mathbf{3}{\mathbf{y}}^2$.

Differentiate the value of M and N.

$\begin{array}{l}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\mathbf{3}{\mathbf{y}}^2\mathbf{+}\mathbf{4}{\mathbf{e}}^x\\ \frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{=}\mathbf{2}{\mathbf{e}}^x\end{array}$

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

$\begin{array}{c}\frac{\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{-}\frac{\partial \mathbf{N}}{\partial \mathbf{x}}}{N}\mathbf{=}\frac{\mathbf{3}{\mathbf{y}}^2\mathbf{+}\mathbf{4}{\mathbf{e}}^x\mathbf{-}\mathbf{2}{\mathbf{e}}^x}{\mathbf{2}{\mathbf{e}}^x\mathbf{+}\mathbf{3}{\mathbf{y}}^2}\\ \mathbf{=}\frac{\mathbf{3}{\mathbf{y}}^2\mathbf{+}\mathbf{2}{\mathbf{e}}^x}{\mathbf{2}{\mathbf{e}}^x\mathbf{+}\mathbf{3}{\mathbf{y}}^2}\\ =1\end{array}$

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

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

Multiply e^x in equation (1) on both sides.

$\begin{array}{l}{\mathbf{e}}^x\left({\mathbf{y}}^3\mathbf{+}\mathbf{4}{\mathbf{e}}^x\mathbf{y}\right)\mathbf{dx}\mathbf{+}{\mathbf{e}}^x\left(\mathbf{2}{\mathbf{e}}^x\mathbf{+}\mathbf{3}{\mathbf{y}}^2\right)\mathbf{dy}\mathbf{=}\mathbf{0}a\\ \left({\mathbf{y}}^3{\mathbf{e}}^x\mathbf{+}\mathbf{4}{\mathbf{e}}^{2x}\mathbf{y}\right)\mathbf{dx}\mathbf{+}\left(\mathbf{2}{\mathbf{e}}^{2x}\mathbf{+}\mathbf{3}{\mathbf{y}}^2{\mathbf{e}}^x\right)\mathbf{dy}\mathbf{=}\mathbf{0}\end{array}$

Now again check the founded equation is exact or not.

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

Differentiate the value of M and N.

$\begin{array}{l}\frac{\partial \mathbf{M}}{\partial \mathbf{y}}\mathbf{=}\mathbf{3}{\mathbf{y}}^2{\mathbf{e}}^x\mathbf{+}\mathbf{4}{\mathbf{e}}^{2x}\\ \frac{\partial \mathbf{N}}{\partial \mathbf{x}}\mathbf{=}\mathbf{4}{\mathbf{e}}^{2x}\mathbf{+}\mathbf{3}{\mathbf{y}}^2{\mathbf{e}}^x\end{array}$

Therefore, the founded equation is exact.

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

Integrate on both sides.

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

Differentiate the F with respect to y.

data-custom-editor="chemistry" $\begin{array}{c}\frac{\partial \mathbf{F}}{\partial \mathbf{y}}\mathbf{=}\mathbf{3}{\mathbf{y}}^2{\mathbf{e}}^x\mathbf{+}\mathbf{2}{\mathbf{e}}^{2x}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(y\right)\\ =N\end{array}$

Equalise the values of N.

$\begin{array}{c}\mathbf{3}{\mathbf{y}}^2{\mathbf{e}}^x\mathbf{+}\mathbf{2}{\mathbf{e}}^{2x}\mathbf{+}\mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}\mathbf{3}{\mathbf{y}}^2{\mathbf{e}}^x\mathbf{+}\mathbf{2}{\mathbf{e}}^{2x}\\ \mathbf{g}\mathbf{\text{'}}\left(y\right)\mathbf{=}0\end{array}$

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 equation of F.

$\begin{array}{c}{\mathbf{y}}^3{\mathbf{e}}^x\mathbf{+}\mathbf{2}{\mathbf{e}}^{2x}\mathbf{y}\mathbf{+}{\mathbf{C}}_1\mathbf{=}\mathbf{0}\\ {\mathbf{y}}^3{\mathbf{e}}^x\mathbf{+}\mathbf{2}{\mathbf{e}}^{2x}\mathbf{y}\mathbf{=}\mathbf{C}\end{array}$

So, the solution is ${\mathbf{y}}^3{\mathbf{e}}^x\mathbf{+}\mathbf{2}{\mathbf{e}}^{2x}\mathbf{y}\mathbf{=}\mathbf{C}$