What is the solution for ty^'+y=t^4y^3 ?

The solution for ty^'+y=t^4y^3 is y= 1/(tsqrt(-t^2+c_{1))},y=-1/(tsqrt(-t^2+c_{1))}

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ty^'+y=t^4y^3

t y^{^{'}} + y = t^{4} y^{3}

y = \frac{1}{t - t ^{2} + c _{1}}, y = - \frac{1}{t - t ^{2} + c _{1}}

Solution steps

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

Solve Bernoulli ODE:

y = \frac{1}{t - t ^{2} + c _{1}}, y = - \frac{1}{t - t ^{2} + c _{1}}

y = \frac{1}{t - t ^{2} + c _{1}}, y = - \frac{1}{t - t ^{2} + c _{1}}

\int (a x^{2} + b x + c) d x

y = \frac{2 x - 1}{x + 2}

\frac{d}{d x} (x^{2} (ln (x)))

\frac{d}{d x} (ln (6 sec (x)))

f (x) = 2 x^{2} - 3 x + 1, 1, 2

What is the solution for ty^'+y=t^4y^3 ?
The solution for ty^'+y=t^4y^3 is y= 1/(tsqrt(-t^2+c_{1))},y=-1/(tsqrt(-t^2+c_{1))}