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Popular >

y^'=xe^{x^2-4}

Solution

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

Solution

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

Solution steps

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

If

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}

Popular Examples

derivative of y=ln(x^3+4)derivative of

y = ln (x^{3} + 4)

(\partial)/(\partial x)(sqrt(6x-2y))

\frac{\partial}{\partial x} (6 x - 2 y)

y^'=4x+y,y(0)=7

y^{^{'}} = 4 x + y, y (0) = 7

y^{''}-6y^'+9y=0,y(0)=2,y^'(0)=1

y^{^{''}} - 6 y^{^{'}} + 9 y = 0, y (0) = 2, y^{^{'}} (0) = 1

derivative of 2/(5x+3)derivative of

\frac{2}{5 x + 3}

Frequently Asked Questions (FAQ)

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