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Popular >

y^'+4y=11,y(0)=1

Solution

y^{^{'}} + 4 y = 11, y (0) = 1

Solution

y = - \frac{7 e ^{- 4 t}}{4} + \frac{11}{4}

Solution steps

y^{'} (t) + 4 y = 11

Solve separable ODE:

y = - \frac{7 e ^{- 4 t}}{4} + \frac{11}{4}

y = - \frac{7 e ^{- 4 t}}{4} + \frac{11}{4}

Graph

Popular Examples

(dx)/(dt)= 4/(xe^{t+6x)}

\frac{d x}{d t} = \frac{4}{x e ^{t + 6 x}}

y^'=((3x^2))/(2y)

y^{^{'}} = \frac{( 3 x ^{2} )}{2 y}

derivative of 4x^3+2y^5=6

\frac{d}{d x} (4 x^{3} + 2 y^{5}) = 6

(dy)/(dx)=(x-1)(y+1)

\frac{d y}{d x} = (x - 1) (y + 1)

t^2y+11ty^'-11y=0,y^1(0)=t

t^{2} y + 11 t y^{^{'}} - 11 y = 0, y^{1} (0) = t

Frequently Asked Questions (FAQ)

What is the solution for y^'+4y=11,y(0)=1 ?
The solution for y^'+4y=11,y(0)=1 is y=-(7e^{-4t})/4+11/4