What is the solution for y^{''}+5ty^'-5y=0 ?

The solution for y^{''}+5ty^'-5y=0 is y=t*integral of (e^{(-5t^2+c_{1})/2})/(t^2)

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y^{''}+5ty^'-5y=0

y^{^{''}} + 5 t y^{^{'}} - 5 y = 0

y = t \cdot \int \frac{e ^{\frac{- 5 t ^{2} + c _{1}}{2}}}{t ^{2}} d t

Solution steps

y^{''} (t) + 5 t y^{'} (t) - 5 y = 0

Solve using substitution:

y = t \cdot \int \frac{e ^{\frac{- 5 t ^{2} + c _{1}}{2}}}{t ^{2}} d t

y = t \cdot \int \frac{e ^{\frac{- 5 t ^{2} + c _{1}}{2}}}{t ^{2}} d t

\frac{d}{d x} (ln (cosh (x)))

\frac{\partial}{\partial y} (x cos (y) + y e^{x})

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

x \to - 2 lim (\frac{2 - x ^{2}}{4 x})

(x^{2} + 1) (\frac{d y}{d x}) + 3 x (y - 1) = 0, y (0) = 6

What is the solution for y^{''}+5ty^'-5y=0 ?
The solution for y^{''}+5ty^'-5y=0 is y=t*integral of (e^{(-5t^2+c_{1})/2})/(t^2)