What is the solution for (dy)/(dx)=xe^{x^2-4} ?

The solution for (dy)/(dx)=xe^{x^2-4} is y= 1/2 e^{x^2-4}+c_{1}

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(dy)/(dx)=xe^{x^2-4}

\frac{d y}{d x} = x e^{x^{2} - 4}

y = \frac{1}{2} e^{x^{2} - 4} + c_{1}

Solution steps

\frac{d y}{d x} = x e^{x^{2} - 4}

Substitute

\frac{d y}{d x}

with

y^{'} (x)

y^{^{'}} (x) = x e^{x^{2} - 4}

f^{^{'}} (x) = g (x)

then

f (x) = \int g (x) d x

y = \int x e^{x^{2} - 4} d x

\int x e^{x^{2} - 4} d x = \frac{1}{2} e^{x^{2} - 4} + c_{1}

y = \frac{1}{2} e^{x^{2} - 4} + c_{1}

n = 1 \sum \infty (- 8)^{- n}

\frac{\partial}{\partial z} (e^{x y} sin (z))

\int (x^{2} + 3 x)^{4} (2 x + 3) d x

4 - 2 e^{- 4 t}

(\frac{ln ( x )}{x})^{^{'}}

What is the solution for (dy)/(dx)=xe^{x^2-4} ?
The solution for (dy)/(dx)=xe^{x^2-4} is y= 1/2 e^{x^2-4}+c_{1}