Answer to Question #261387 in Differential Equations for Dionney

Question

Solve the differential equation by substitution suggested by equation. Show complete solution.

(5x+3e^y)dx + 2xe^y dy =0

Answer 1

We do substitution $u=e^y,du=e^ydy$ and have

(5x+3u)dx+2xdu=0;

$2x\frac{du}{dx}+5x+3u=0;\\ u'(x)+\frac{5}{2}+\frac{3}{2}\frac{u}{x}=0; t=\frac{u}{x}- new \space variable;\\ u'(x)=t'(x)x+t;\\ t'+\frac{5}{2}+\frac{5}{2}\cdot t=0;\\$

$\frac{2}{5}\frac{dt}{t+1}=-dx;$

$\frac{2}{5}\int \frac{dt}{t+1}=-\int dx=-x+C;\\ ln|t+1|=C-\frac{5}{2}x;\\ t+1=\pm e^C\cdot e^{-\frac{5}{2}x}=C\cdot e^{-\frac{5}{2}x},where \space C:=\pm e^C$

$t=-1+C\cdot e^{-\frac{5}{2}x};\\ \frac{u}{x}=-1+C\cdot e^{-\frac{5}{2}x};\\ u=-x+C\cdot x\cdot e^{-\frac{5}{2}x};\\$ $C\in R$

$y=ln(u)=ln \left(x(C\cdot e^{-\frac{5}{2}x}-1) \right)$ - general solution