What is the solution for t^2y^'+y=sin(t) ?

The solution for t^2y^'+y=sin(t) is y=e^{1/t}*integral of (e^{-1/t}sin(t))/(t^2)

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t^2y^'+y=sin(t)

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

y = e^{\frac{1}{t}} \cdot \int \frac{e ^{- \frac{1}{t}} sin ( t )}{t ^{2}} d t

Solution steps

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

Solve linear ODE:

y = e^{\frac{1}{t}} \cdot \int \frac{e ^{- \frac{1}{t}} sin ( t )}{t ^{2}} d t

y = e^{\frac{1}{t}} \cdot \int \frac{e ^{- \frac{1}{t}} sin ( t )}{t ^{2}} d t

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

y = x^{2}, at x = - \frac{3}{4}

\frac{d}{d x} (ln ∣ csc (x) - cot (x) ∣)

y = x^{2} - 2 x, y = 2 - x

\int 3^{4 x + 6} d x

What is the solution for t^2y^'+y=sin(t) ?
The solution for t^2y^'+y=sin(t) is y=e^{1/t}*integral of (e^{-1/t}sin(t))/(t^2)