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Popular >

y^'+y=te^t

Solution

y^{^{'}} + y = t e^{t}

Solution

y = \frac{e ^{t} ( 2 t - 1 )}{4} + \frac{c _{1}}{e ^{t}}

Solution steps

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

Solve linear ODE:

y = \frac{e ^{t} ( 2 t - 1 )}{4} + \frac{c _{1}}{e ^{t}}

y = \frac{e ^{t} ( 2 t - 1 )}{4} + \frac{c _{1}}{e ^{t}}

Popular Examples

x(dy)/(dx)+4y=2-7/x

x \frac{d y}{d x} + 4 y = 2 - \frac{7}{x}

integral of (x^3+4x-3)

\int (x^{3} + 4 x - 3) d x

derivative of 5x^{-15}derivative of

5 x^{- 15}

integral of (10x+15)/(x^2+3x)

\int \frac{10 x + 15}{x ^{2} + 3 x} d x

(dy)/(dx)=(2-x)/(y-3)

\frac{d y}{d x} = \frac{2 - x}{y - 3}

Frequently Asked Questions (FAQ)

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