Answer to Question #261373 in Differential Equations for Dionne

Question

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

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

Answer 1

$(5x+3e^y)dx+2xe^y dy =0\\ 2xe^y dy=-((5x+3e^y)dx)\\ e^y \frac{dy}{dx}=-(\frac{5x+3e^y}{2x})$

Let us substitute $e^y=z\Rightarrow e^y \frac{dy}{dx}=\frac{dz}{dx}\\$

Then, $\frac{dz}{dx}=-(\frac{5x+3e^y}{2x})$

$\Rightarrow \frac{d z}{d x}=-\frac{5}{2}-\frac{3}{2 x} \cdot z \\ \Rightarrow \frac{d z}{d x}+\frac{3}{2 x} \cdot z=-\frac{5}{2}$

$I \cdot F=e^{\int \frac{3}{2 x} d x}=e^{\frac{3}{2} \cdot \ln x}=e^{\ln x^{3 / 2}}=x^{3 / 2}$

Multiplying above differential equation by I . F, and integrating,

$\Rightarrow z \cdot x^{3 / 2}=\int -\frac{5}{2}.x^\frac{3}{2}dx+c\\ \Rightarrow z \cdot x^{3 / 2}=-\frac{5}{2} \cdot \frac{x^{\frac{3}{2}+1}}{\frac{3}{2}+1}+c \\ \Rightarrow z \cdot x^{3 / 2}=-\frac{5}{2} \cdot \frac{2}{5} \cdot x^{\frac{5}{2}}+c\\ \Rightarrow e^{y} \cdot x^{3 / 2}=-x^{\frac{5}{2}}+c\left[\because z=e^{y}\right]\\$

$\Rightarrow e^{y}=-x^{\frac{5}{2}-\frac{3}{2}}+c. x^{-3 / 2}\\ \Rightarrow e^{y}=-x+c x^{-\frac{3}{2}}\\ \Rightarrow e^{y}=c x^{-\frac{3}{2}}-x\\ \Rightarrow y=ln|c x^{-\frac{3}{2}}-x|$