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Popular >

t^2y^'-t=1+y+ty

Solution

t^{2} y^{^{'}} - t = 1 + y + t y

Solution

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

Solution steps

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

Solve separable ODE:

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

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

Popular Examples

integral of (x+5)/(x^2+4x+8)

\int \frac{x + 5}{x ^{2} + 4 x + 8} d x

derivative of f(x)=e^{3sqrt(x)}derivative of

f (x) = e^{3 x}

(tdy)/(dt)=y+5t^2sin(t)

\frac{t d y}{d t} = y + 5 t^{2} sin (t)

derivative of (x-2^3(x^2+9)^4)

\frac{d}{d x} ((x - 2)^{3} (x^{2} + 9)^{4})

derivative of (3a^2/(x^2)-a/(3x^4))

\frac{d}{d x} (\frac{3 a ^{2}}{x ^{2}} - \frac{a}{3 x ^{4}})

Frequently Asked Questions (FAQ)

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