Question

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

$\mathbf{35}\mathbf{.}\mathbf{}\left(10-6y+e^{-3x}\right)\mathit{d}\mathit{x}\mathbf{-}\mathbf{2}\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}$

Short answer

The resulting equation is$-2e^{3x}y+\frac{10}{3}{e}^{3x}+x=c$

Step-by-step solution

Identifying M and N from differential equation

$$\begin{gathered} M=10-6y+e^{-3x} \\ N=-2 \\ Finding{N}_{x}and{M}_y \\ {M}_y=-6 \\ {N}_x=0 \\ \frac{M_y-N_x}{N}=\frac{-6-0}{-2}=3Compute\frac{M_y-N_x}{N} \\ \mu \left(x\right)=e^{\int 3dx}=e^{3x} \end{gathered}$$

Since $\frac{M_y-N_x}{N}$only depend on x, The integrating factor$=e^{\int \frac{M_y-N_x}{N}dx}$

$\left(10e^{3x}-6e^{3x}y+1\right)dx-2e^{3x}dy=0$

Multiply both sides of the differentia equation by the integrating factor

$$\begin{gathered} M\left(x,y\right)=10e^{3x}-6e^{3x}y+1 \\ \leftrightarrow \frac{\partial M}{\partial y}=-6e^{3x} \\ N\left(x,y\right)=-2e^{3x} \\ \leftrightarrow \frac{\partial N}{\partial x}=-6e^{3x} \end{gathered}$$

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

$$\begin{gathered} f\left(x,y\right)=-2e^{3x}y+g\left(x\right) \\ IntegrateN\left(x,y\right)=\frac{\partial f}{\partial y}=-2e^{3x} \\ \frac{\partial f}{\partial x}=-6e^{3x}+g\text{'}\left(x\right) \\ Take\frac{\partial f}{\partial y} \\ g\text{'}\left(x\right)=10e^{3x}+1 \\ g\left(x\right)=\frac{10}{3}{e}^{3x}+x \end{gathered}$$

Final proof

Since $M\left(x,y\right)=\frac{\partial f}{\partial y}$, we see that$g\text{'}\left(x\right)=10e^{3x}+1$

Final solution$-2e^{3x}y+\frac{10}{3}{e}^{3x}+x=c$