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Popular >

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

Solution

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

Solution

y = - x - 1 + e^{x}

Solution steps

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

Solve linear ODE:

y = - x - 1 + e^{x}

y = - x - 1 + e^{x}

Popular Examples

integral of u^2

\int u^{2} d u

integral from 0 to 3 of sqrt(3x)

\int_{0}^{3} 3 x d x

derivative of 2-3x^2

\frac{d}{d x} (2 - 3 x^{2})

(\partial)/(\partial y)(x^{3/4}y^{1/4})

\frac{\partial}{\partial y} (x^{\frac{3}{4}} y^{\frac{1}{4}})

area y=x^2,y=4 area

y = x^{2}, y = 4

Frequently Asked Questions (FAQ)

What is the solution for y^'=x+y,y(0)=0 ?
The solution for y^'=x+y,y(0)=0 is y=-x-1+e^x