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Popular Calculus >

10y^'+20y=12

Solution

10 y^{^{'}} + 20 y = 12

Solution

y = - \frac{e ^{- 2 t - c_{1}}}{5} + \frac{3}{5}

Solution steps

10 y^{'} (t) + 20 y = 12

Solve separable ODE:

y = - \frac{e ^{- 2 t - c_{1}}}{5} + \frac{3}{5}

y = - \frac{e ^{- 2 t - c_{1}}}{5} + \frac{3}{5}

Graph

Popular Examples

2(12)(18)+2(9)(dy)/(dt)=0

2 (12) (18) + 2 (9) \frac{d y}{d t} = 0

(dy)/(dt)=25t^4

\frac{d y}{d t} = 25 t^{4}

y^'=(6x-12)/(x^2),y(1)=20

y^{^{'}} = \frac{6 x - 12}{x ^{2}}, y (1) = 20

integral of 2xln(x^2)

\int 2 x ln (x^{2}) d x

integral of (cos^3(x))/(sin^2(x))

\int \frac{cos ^{3} ( x )}{sin ^{2} ( x )} d x

Frequently Asked Questions (FAQ)

What is the solution for 10y^'+20y=12 ?
The solution for 10y^'+20y=12 is y=-(e^{-2t-c_{1}})/5+3/5