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Popular >

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

Solution

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

Solution

y = - x - 1 + 6 e^{x}

Solution steps

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

Solve linear ODE:

y = - x - 1 + 6 e^{x}

y = - x - 1 + 6 e^{x}

Popular Examples

integral of x/((x-2)^2)

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

limit as x approaches-infinity of x-1

x \to - \infty lim (x - 1)

(\partial ^2)/(\partial x^2)(xy^2-6x^2-3y^2)

\frac{\partial ^{2}}{\partial x ^{2}} (x y^{2} - 6 x^{2} - 3 y^{2})

derivative of (2x^2+x-3)/(-3x^2-4x+2)derivative of

\frac{2 x ^{2} + x - 3}{- 3 x ^{2} - 4 x + 2}

simplify (e^{-x}-e^5)(e^x+e^{-3})simplify

(e^{- x} - e^{5}) (e^{x} + e^{- 3})

Frequently Asked Questions (FAQ)

What is the solution for y^'=x+y,y(0)=5 ?
The solution for y^'=x+y,y(0)=5 is y=-x-1+6e^x