Question

In Problems 31-36 solve the given differential equation by finding as in Example 4, an appropriate integrating factor.

$\mathbf{36}\mathbf{.}\mathbf{}\left(y^2+x{y}^3\right)\mathit{d}\mathit{x}\mathbf{+}\left(5y^2-xy+y^{3}siny\right)\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The resulting equation is$5\ln y+\frac{x}{y}-\cos y+\frac{1}{2}{x}^2=c$

Step-by-step solution

Identifying M and N from differential equation

$$\begin{gathered} M=y^2+x{y}^3 \\ N=5y^2-xy+y^3\sin y \\ Finding{N}_{x}and{M}_y \\ {M}_y=2y+3x{y}^2 \\ {N}_x=-y \\ \frac{M_y{N}_x}{N}=\frac{2y+3x{y}^2-\left(-y\right)}{5y^2-xy+y^3\sin y}=\frac{3y+3x{y}^2}{5y^2-xy+y^3\sin y}Compute\frac{M_y{N}_x}{N} \\ \frac{N_x-M_y}{M}=\frac{-y-2y-3x{y}^2}{y^2+x{y}^3}=\frac{-3y-3x{y}^2}{y^2+x{y}^3} \\ \frac{-3y\left(1+xy\right)}{y^2\left(1+xy\right)}=-\frac{3}{y}Compute\frac{N_x-M_y}{M} \\ \mu \left(y\right)=e^{\int -\frac{3}{y}dy}=e^{-3\ln y}=e^{\ln y^{-3}}=y^{-3} \end{gathered}$$

Since $\frac{N_x-M_y}{M}$only depends on x, the integrating factor $=e^{\int \frac{N_x-M_y}{M}}dy$

$\left(\frac{1}{y}+x\right)dx+\left(\frac{5}{y}-\frac{x}{y^2}+\sin y\right)dy=0$

Multiply both sides of the differentia equation by the integrating factor

$$\begin{gathered} M\left(x,y\right)=\frac{1}{y}+x \\ \leftrightarrow \frac{\partial M}{\partial y}=-\frac{1}{y^2} \\ N\left(x,y\right)=\frac{5}{y}-\frac{x}{y^2}+\sin y \\ \leftrightarrow \frac{\partial N}{\partial x}=-\frac{1}{y^2} \end{gathered}$$

Checking whether the differential equation

is exact

$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

Therefore, the equation is exact

Find the integrate

$$\begin{gathered} f\left(x,y\right)=5\ln y+\frac{x}{y}-\cos y+g\left(x\right) \\ IntegrateN\left(x,y\right)=\frac{\partial f}{\partial y}=\frac{5}{y}-\frac{x}{y^2}+\sin y \\ \frac{\partial f}{\partial x}=\frac{5}{y}+g\text{'}\left(x\right) \\ Take\frac{\partial f}{\partial x} \\ g\text{'}\left(x\right)=x \\ g\left(x\right)=\frac{1}{2}{x}^2 \end{gathered}$$

Final proof

Since $M\left(x,y\right)=\frac{\partial f}{\partial y}$, we see that

g'(x) = x

Final solution$5\ln y+\frac{x}{y}-\cos y+\frac{1}{2}{x}^2=c$