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Popular >

y^'=x+y,y(0)=7

Solution

y^{^{'}} = x + y, y (0) = 7

Solution

y = - x - 1 + 8 e^{x}

Solution steps

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

Solve linear ODE:

y = - x - 1 + 8 e^{x}

y = - x - 1 + 8 e^{x}

Popular Examples

derivative of (4x^2+2x+8/(sqrt(x)))

\frac{d}{d x} (\frac{4 x ^{2} + 2 x + 8}{x})

derivative of 2x^{-2}

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

y^{''}-8y^'+20y=0

y^{^{''}} - 8 y^{^{'}} + 20 y = 0

limit as x approaches pi of e^{cos(3x)}

x \to π lim (e^{c o s (3 x)})

integral from 1 to sqrt(2 of)x5^{(x^2)}

\int_{1}^{2} x 5^{(x^{2})} d x

Frequently Asked Questions (FAQ)

What is the solution for y^'=x+y,y(0)=7 ?
The solution for y^'=x+y,y(0)=7 is y=-x-1+8e^x