Answer to Question #300202 in Differential Equations for Joshua

Question

Use appropriate substitution to reduce the following equation to a variable separable and then solve the given IVP: $dy/dx= (3x+2y)/(3x+2y+2), y(-1)=-1$

Answer 1

Let us solve the differential equation

$\frac{dy}{dx}=\frac{3x+2y}{3x+2y+2},\ \ y(−1)=−1.$

Let us use the transformation $u=3x+2y.$ Then $\frac{du}{dx}=3+2\frac{dy}{dx},$ and hence $\frac{dy}{dx}=\frac{1}2(\frac{du}{dx}-3).$

It follows that

$\frac{1}2(\frac{du}{dx}-3)=\frac{u}{u+2},$ and hence $\frac{du}{dx}=\frac{2u}{u+2}+3=\frac{5u+6}{u+2}.$

We get the equation ${dx}=\frac{(u+2)du}{5u+6},$ and hence $\int{dx}=\int\frac{(u+2)du}{5u+6}=\frac{1}5\int\frac{(5u+10)du}{5u+6} =\frac{1}5\int(1+\frac{4}{5u+6})du=\frac{u}5+\frac{4}{25}\ln|5u+6|+C.$

Therefore,

$x=\frac{3x+2y}5+\frac{4}{25}\ln|5(3x+2y)+6|+C$

Since $y(-1)=-1,$ we conclude that $-1=-1+\frac{4}{25}\ln 19+C,$ and thus $C=-\frac{4}{25}\ln 19.$

We conclude that the solution of $\frac{dy}{dx}=\frac{3x+2y}{3x+2y+2},\ \ y(−1)=−1$ is

$x=\frac{3x+2y}5+\frac{4}{25}\ln|15x+10y+6|-\frac{4}{25}\ln 19.$