What is the solution for y^'+3x^2y=sin(x)e^{-x^3},y(0)=1 ?

The solution for y^'+3x^2y=sin(x)e^{-x^3},y(0)=1 is y=-(cos(x))/(e^{x^3)}+2/(e^{x^3)}

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y^'+3x^2y=sin(x)e^{-x^3},y(0)=1

y^{^{'}} + 3 x^{2} y = sin (x) e^{- x^{3}}, y (0) = 1

y = - \frac{cos ( x )}{e ^{x^{3}}} + \frac{2}{e ^{x^{3}}}

Solution steps

y^{'} (x) + 3 x^{2} y = sin (x) e^{- x^{3}}

Solve linear ODE:

y = - \frac{cos ( x )}{e ^{x^{3}}} + \frac{2}{e ^{x^{3}}}

y = - \frac{cos ( x )}{e ^{x^{3}}} + \frac{2}{e ^{x^{3}}}

x^{3} b^{5} + y^{4} b + z^{6}

\int - sin (t) cos (t) d t

\int \frac{ln ( x ) + 9}{x} d x

\frac{\partial}{\partial x} (2 x^{2} + y)

tan (2 (x + 1))

What is the solution for y^'+3x^2y=sin(x)e^{-x^3},y(0)=1 ?
The solution for y^'+3x^2y=sin(x)e^{-x^3},y(0)=1 is y=-(cos(x))/(e^{x^3)}+2/(e^{x^3)}