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Popular >

y^'+100y=5

Solution

y^{^{'}} + 100 y = 5

Solution

y = - \frac{e ^{- 100 t - c_{1}}}{20} + \frac{1}{20}

Solution steps

y^{'} (t) + 100 y = 5

Solve separable ODE:

y = - \frac{e ^{- 100 t - c_{1}}}{20} + \frac{1}{20}

y = - \frac{e ^{- 100 t - c_{1}}}{20} + \frac{1}{20}

Popular Examples

(dy)/(dx)+6x^2y=5x^2,y(0)=2

\frac{d y}{d x} + 6 x^{2} y = 5 x^{2}, y (0) = 2

(dy)/(dx)=sin(x)(1-cos(x))(y^2-4)

\frac{d y}{d x} = sin (x) (1 - cos (x)) (y^{2} - 4)

(dy)/(dx)=((x^2+1))/(2-2y)

\frac{d y}{d x} = \frac{( x ^{2} + 1 )}{2 - 2 y}

(x^2+1)tan(y)y^'=x

(x^{2} + 1) tan (y) y^{^{'}} = x

(dy)/(dt)=(6t^2)/(7y)

\frac{d y}{d t} = \frac{6 t ^{2}}{7 y}

Frequently Asked Questions (FAQ)

What is the solution for y^'+100y=5 ?
The solution for y^'+100y=5 is y=-(e^{-100t-c_{1}})/(20)+1/20