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Popular >

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

Solution

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

Solution

y = - x - 1 + 2 e^{x}

Solution steps

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

Solve linear ODE:

y = - x - 1 + 2 e^{x}

y = - x - 1 + 2 e^{x}

Popular Examples

(dy)/(dx)=(y^2-1)/(x^2-1),y(2)=2

\frac{d y}{d x} = \frac{y ^{2} - 1}{x ^{2} - 1}, y (2) = 2

integral of sin(t)e^{-2/3 t}

\int sin (t) e^{- \frac{2}{3} t} d t

integral from 1 to 2 of x^2e^{x^3-1}

\int_{1}^{2} x^{2} e^{x^{3} - 1} d x

sum from n=1 to infinity of 1/(\sqrt[n]{2)}

y^{''}+9y=(sec(3t))^2

y^{^{''}} + 9 y = (sec (3 t))^{2}

Frequently Asked Questions (FAQ)

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