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Popular >

y^'=4x+y,y(0)=1

Solution

y^{^{'}} = 4 x + y, y (0) = 1

Solution

y = - 4 (x + 1) + 5 e^{x}

Solution steps

y^{'} (x) = 4 x + y

Solve linear ODE:

y = - 4 (x + 1) + 5 e^{x}

y = - 4 (x + 1) + 5 e^{x}

Popular Examples

integral from 0 to 6 of sqrt(6x)-(3x-12)

\int_{0}^{6} 6 x - (3 x - 12) d x

integral from 0 to infinity of 1/(7+x)

\int_{0}^{\infty} \frac{1}{7 + x} d x

integral of e^2x

\int e^{2} x d x

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

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

integral of x^{-4}+2x+3

\int x^{- 4} + 2 x + 3 d x

Frequently Asked Questions (FAQ)

What is the solution for y^'=4x+y,y(0)=1 ?
The solution for y^'=4x+y,y(0)=1 is y=-4(x+1)+5e^x