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Popular Calculus >

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}

Graph

Popular Examples

derivative of 2x^{-2}

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

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

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

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

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

limit as x approaches 0 of (3x^2-2x)/x

x \to 0 lim (\frac{3 x ^{2} - 2 x}{x})

integral of (e^{2x}-6e^x)/(e^{2x)+11}

\int \frac{e ^{2 x} - 6 e ^{x}}{e ^{2 x} + 11} 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