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Popular >

y^'=2xy+x,y(0)=1

Solution

y^{^{'}} = 2 x y + x, y (0) = 1

Solution

y = \frac{3 e ^{x^{2}}}{2} - \frac{1}{2}

Solution steps

y^{'} (x) = 2 x y + x

Solve separable ODE:

y = \frac{3 e ^{x^{2}}}{2} - \frac{1}{2}

y = \frac{3 e ^{x^{2}}}{2} - \frac{1}{2}

Graph

Popular Examples

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(1/y \frac{dy)/(dx)}{(-1+y)}=-x

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(x+xy^2)dx-4ydy=0,y(1)=0

(x + x y^{2}) d x - 4 y d y = 0, y (1) = 0

y^'+ycos(x)=7cos(x)

y^{^{'}} + y cos (x) = 7 cos (x)

Frequently Asked Questions (FAQ)

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