What is the solution for 2txdx+(t^2-x^2)dt=0 ?

The solution for 2txdx+(t^2-x^2)dt=0 is x=sqrt(-t^2+c_{1)t},x=-sqrt(-t^2+c_{1)t}

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2txdx+(t^2-x^2)dt=0

2 t x d x + (t^{2} - x^{2}) d t = 0

x = - t^{2} + c_{1} t, x = - - t^{2} + c_{1} t

Solution steps

2 t x d x + (t^{2} - x^{2}) d t = 0

Solve Bernoulli ODE:

x = - t^{2} + c_{1} t, x = - - t^{2} + c_{1} t

x = - t^{2} + c_{1} t, x = - - t^{2} + c_{1} t

x \to 3 - lim (x^{2} + 2 x)

\frac{d}{d x} (8 x^{7})

\int tan (X) d X

y = \frac{6}{x}

\frac{d}{d x} (4 x x)

What is the solution for 2txdx+(t^2-x^2)dt=0 ?
The solution for 2txdx+(t^2-x^2)dt=0 is x=sqrt(-t^2+c_{1)t},x=-sqrt(-t^2+c_{1)t}