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Popular >

y^'=8x+y,y(0)=1

Solution

y^{^{'}} = 8 x + y, y (0) = 1

Solution

y = - 8 (x + 1) + 9 e^{x}

Solution steps

y^{'} (x) = 8 x + y

Solve linear ODE:

y = - 8 (x + 1) + 9 e^{x}

y = - 8 (x + 1) + 9 e^{x}

Graph

Popular Examples

(dy)/(dx)=(2xy-y^2)/(x^2)

\frac{d y}{d x} = \frac{2 x y - y ^{2}}{x ^{2}}

(dy)/(dx)=(4x^3)/(cos(y))

\frac{d y}{d x} = \frac{4 x ^{3}}{cos ( y )}

derivative of (2x-1(3x+5))

\frac{d}{d x} ((2 x - 1) (3 x + 5))

limit as x approaches 2-of x^2+2

x \to 2 - lim (x^{2} + 2)

integral of x^3-6x+7

\int x^{3} - 6 x + 7 d x

Frequently Asked Questions (FAQ)

What is the solution for y^'=8x+y,y(0)=1 ?
The solution for y^'=8x+y,y(0)=1 is y=-8(x+1)+9e^x