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Answer to Question #17052 in Differential Equations for arif

Question #17052
find the solution of separable ODE
dy+x^2dx=x^2e^ydx
Expert's answer
we can substitute
z=y+x^3/3
then we have

dz=dy+x^2*dx
dy=dz-x^2*dx
dy+x^2dx=x^2e^ydx
dz-x^2*dx+x^2*dx=x^2*e^(z-x^3/3)dx
dz=x^2*e^z*e^(-x^3/3)dx
e^(-z)dz=x^2*e^(-x^3/3)dx
-e^(-z)=-c-e^(-x^3/3)
e^(-z)=c+e^(-x^3/3)
-z=ln(c+e^(-x^3/3))
z=-ln(c+e^(-x^3/3))
y=-ln(c+e^(-x^3/3))-x^3/3

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