Answer to Question #237929 in Mechanical Engineering for Hercu

Question #237929
y' - 2y = y^2 ; y = 3 when x = 0
1
Expert's answer
2021-09-17T02:39:02-0400

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=e2x+Cy+2e2x+C

y(1-2e2x+C)=2e2x+C

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

Substitute y=3 and x=0

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

2eC=3-6eC

8eC=3

eC="\\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}}"


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