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

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

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y^{''}-2/(t^2)y=0,y(-1)=-1,y^'(-1)=-4

y^{^{''}} - \frac{2}{t ^{2}} y = 0, y (- 1) = - 1, y^{^{'}} (- 1) = - 4

y = t^{2} + \frac{2}{t}

Solution steps

y^{''} (t) - \frac{2}{t ^{2}} y = 0

Solve linear ODE:

y = t^{2} + \frac{2}{t}

y = t^{2} + \frac{2}{t}

\int \frac{2 x ^{2} + x + 1}{( x + 3 ) ( x - 1 ) ^{2}} d x

n = 1 \sum \infty 1 6^{n} x^{n} n!

\frac{d y}{d x} = 6 x - 30

\int x x^{2} + 7 d x

\int \frac{1}{2000 ( 2000 - x )} d x

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