Question
(5) Solve the following initial value problem:
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(y^2/x^2) dy/dx = e^(2x+3y) , y(0)=0
Image: HT TP s : / / ibb.co / 3k9L2sw
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(y^2/x^2) dy/dx = e^(2x+3y) , y(0)=0
Image: HT TP s : / / ibb.co / 3k9L2sw
Answers
oobleck
(y^2/x^2) y' = e^(2x+3y) = e^(2x)*e^(3y)
y^2 e^(-3y) dy = x^2 e^(2x) dx
-1/27 (9y^2 + 6y + 2) e^(-3y) = 1/4 (2x^2 - 2x + 1) e^(2x) + C
Since y(0) = 0, we have
-1/27 (2) = 1/4 (1) + C
C = -35/108
and you wind up with
-1/27 (9y^2 + 6y + 2) e^(-3y) = 1/4 (2x^2 - 2x + 1) e^(2x) - 35/108
y^2 e^(-3y) dy = x^2 e^(2x) dx
-1/27 (9y^2 + 6y + 2) e^(-3y) = 1/4 (2x^2 - 2x + 1) e^(2x) + C
Since y(0) = 0, we have
-1/27 (2) = 1/4 (1) + C
C = -35/108
and you wind up with
-1/27 (9y^2 + 6y + 2) e^(-3y) = 1/4 (2x^2 - 2x + 1) e^(2x) - 35/108