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

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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