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Popular >

y^'+4y=12,y(0)=0

Solution

y^{^{'}} + 4 y = 12, y (0) = 0

Solution

y = - 3 e^{- 4 t} + 3

Solution steps

y^{'} (t) + 4 y = 12

Solve separable ODE:

y = - 3 e^{- 4 t} + 3

y = - 3 e^{- 4 t} + 3

Popular Examples

(dy)/(dx)+yln(x)=3ln(x)

\frac{d y}{d x} + y ln (x) = 3 ln (x)

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

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

y^'=(1+y^6)/(xy^5),y(1)=1

y^{^{'}} = \frac{1 + y ^{6}}{x y ^{5}}, y (1) = 1

(dy)/(dx)ax^2y=7

\frac{d y}{d x} a x^{2} y = 7

(dz)/(dt)=6te^{4z}

\frac{d z}{d t} = 6 t e^{4 z}

Frequently Asked Questions (FAQ)

What is the solution for y^'+4y=12,y(0)=0 ?
The solution for y^'+4y=12,y(0)=0 is y=-3e^{-4t}+3