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Popular >

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

Solution

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

Solution

y = \frac{2 t - 1}{4} + \frac{5}{4 e ^{2 t}}

Solution steps

y^{'} (t) + 2 y = t

Solve linear ODE:

y = \frac{2 t - 1}{4} + \frac{5}{4 e ^{2 t}}

y = \frac{2 t - 1}{4} + \frac{5}{4 e ^{2 t}}

Popular Examples

(dy}{dx}-2y=\frac{2y)/x

\frac{d y}{d x} - 2 y = \frac{2 y}{x}

integral of arccot(10y)

\int arccot (10 y) d y

derivative of ((x-5(x^2+5x+25))/(x^3))

\frac{d}{d x} (\frac{( x - 5 ) ( x ^{2} + 5 x + 25 )}{x ^{3}})

derivative of 4x^2+4x

\frac{d}{d x} (4 x^{2} + 4 x)

integral of (x+2)/(2x+1)

\int \frac{x + 2}{2 x + 1} d x

Frequently Asked Questions (FAQ)

What is the solution for y^'+2y=t,y(0)=1 ?
The solution for y^'+2y=t,y(0)=1 is y=(2t-1)/4+5/(4e^{2t)}