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Popular >

y^'+4y=32x,y(0)=3

Solution

y^{^{'}} + 4 y = 32 x, y (0) = 3

Solution

y = 2 (4 x - 1) + \frac{5}{e ^{4 x}}

Solution steps

y^{'} (x) + 4 y = 32 x

Solve linear ODE:

y = 2 (4 x - 1) + \frac{5}{e ^{4 x}}

y = 2 (4 x - 1) + \frac{5}{e ^{4 x}}

Graph

Popular Examples

(dy)/(dx)=xy+5x+4y+20

\frac{d y}{d x} = x y + 5 x + 4 y + 20

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\frac{d}{d x} (7 e^{- 5 x})

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\frac{d}{d x} (\frac{10}{e ^{x}})

tangent of f(x)=2x^4,\at x=-2,32 tangent of

f (x) = 2 x^{4}, at x = - 2, 32

derivative of (6x^2+6x+8)/(sqrt(x))derivative of

\frac{6 x ^{2} + 6 x + 8}{x}

Frequently Asked Questions (FAQ)

What is the solution for y^'+4y=32x,y(0)=3 ?
The solution for y^'+4y=32x,y(0)=3 is y=2(4x-1)+5/(e^{4x)}