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Popular >

y^'=2x+y,y(0)=5

Solution

y^{^{'}} = 2 x + y, y (0) = 5

Solution

y = - 2 (x + 1) + 7 e^{x}

Solution steps

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

Solve linear ODE:

y = - 2 (x + 1) + 7 e^{x}

y = - 2 (x + 1) + 7 e^{x}

Popular Examples

derivative of-e^x

\frac{d}{d x} (- e^{x})

integral of cot^4(x)csc^4(x)

\int cot^{4} (x) csc^{4} (x) d x

integral from 0 to pi/2 of 5cos^2(x)

\int_{0}^{\frac{π}{2}} 5 cos^{2} (x) d x

y^{''}-4y^'+16y=0

y^{^{''}} - 4 y^{^{'}} + 16 y = 0

(\partial)/(\partial x)(1/(x+2ct))

\frac{\partial}{\partial x} (\frac{1}{x + 2 c t})

Frequently Asked Questions (FAQ)

What is the solution for y^'=2x+y,y(0)=5 ?
The solution for y^'=2x+y,y(0)=5 is y=-2(x+1)+7e^x