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Jake
solve each differential equation, applying the initial condition when provided. dy/dx=e^x/2y and dy/dx=3y y(0)=5
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Answered by
John
I would also like to see the answer to this one!
Answered by
oobleck
dy/dx=e^x/2y
2y dy = e^x dx
y^2 = e^x + c
dy/dx = 3y
dy/y = 3 dx
ln y = 3x+c
y = ce^(3x)
y(0)=5, so c=5
y = 5e^(3x)
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