What is the solution for t^2y^'+4ty-y^3=0,y(1)= 3/(sqrt(2)) ?

The solution for t^2y^'+4ty-y^3=0,y(1)= 3/(sqrt(2)) is y=(3sqrt(t))/(sqrt(2))

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t^2y^'+4ty-y^3=0,y(1)= 3/(sqrt(2))

t^{2} y^{^{'}} + 4 t y - y^{3} = 0, y (1) = \frac{3}{2}

y = \frac{3 t}{2}

Solution steps

t^{2} y^{'} (t) + 4 t y - y^{3} = 0

Solve Bernoulli ODE:

y = \frac{3 t}{2}

y = \frac{3 t}{2}

f (x) = x 2^{x}

\frac{d}{d y} (y^{- 3})

\int_{0}^{3} (2 sin (x) - e^{x}) d x

n = 1 \sum \infty \frac{1}{n} \cdot (- 1)^{n}

\int \frac{sin ( 2 x )}{10 + cos ^{2} ( x )} d x

What is the solution for t^2y^'+4ty-y^3=0,y(1)= 3/(sqrt(2)) ?
The solution for t^2y^'+4ty-y^3=0,y(1)= 3/(sqrt(2)) is y=(3sqrt(t))/(sqrt(2))