Answer to Question #237929 in Mechanical Engineering for Hercu

Question

y' - 2y = y^2 ; y = 3 when x = 0

Answer 1

y'-2y=y²

Rewriting the equation,

y''=y²+2y

$\frac{dy}{dx}=y^{2}+2y$

$\frac{dy}{y^{2}+2y}=dx$

$\frac{dy}{(y^{2}+2y+1)-1}=dx$

$\frac{dy}{(y+1)^{2}-1}=dx$

Integrate both sides

$\int \frac{dy}{(y+1)^{2}-1}=\int dx$

$\frac{1}{2}ln|\frac{y}{y+2}|=x+C$

$\frac{y}{y+2}=e^{2x+C}$

y=e^2x+Cy+2e^2x+C

y(1-2e^2x+C)=2e^2x+C

y=$\frac{2e^{2x+C}}{1-2e^{2x+C}}$

Substitute y=3 and x=0

3=$\frac{2e^{C}}{1-2e^{C}}$

2e^C=3-6e^C

8e^C=3

e^C=$\frac{3}{8}$

Hence,

y=$\frac{2*\frac{3}{8}e^{2x}}{1-2*\frac{3}{8}e^{2x}}$

=$\frac{\frac{3}{4}e^{2x}}{1-\frac{3}{4}e^{2x}}$

Thus,y=$\frac{3e^{2x}}{4-3e^{2x}}$