Answer
$y=\frac{1}{6}(\ln(\frac{5}{2}-\frac{3}{2}e^{-4x})$
Work Step by Step
We are given:
$$e^{-3x+2y}dx+e^{x-4y}dy=0$$
$$\frac{dy}{dx}=\frac{e^{-3x+2y}}{e^{x-4y}}$$
$$\frac{dy}{dx}=-e^{6y}e^{-4x}$$
Rewrite as:
$$e^{-6y}dy=e^{-4x}dx$$
Integrating both sides:
$$-\frac{1}{6}e^{-6y}=\frac{1}{4}e^{-4x}+C$$
$C$ is a integration constant
$$-2e^{-6y}=3e^{-4x}+C_1$$
$$e^{-6y}=-\frac{3}{2}e^{-4x}+C_2$$
$$-6y=\ln(-\frac{3}{2}e^{-4x}+C_2)$$
The general solution is $y=\frac{1}{6}(\ln(C_2-\frac{3}{2}e^{-4x})$
We are given $y(0)=0$
Substituting:
$$\frac{1}{6}(\ln(C_2-\frac{3}{2}e^{0})=0$$
$$\ln(C_2-\frac{3}{2})=0$$
$$C=\frac{5}{2}$$
The final solution is:
$$y=\frac{1}{6}(\ln(\frac{5}{2}-\frac{3}{2}e^{-4x})$$
