What is the solution for (25+x^2)y^'-2xy=0 ?

The solution for (25+x^2)y^'-2xy=0 is y=25e^{c_{1}}+e^{c_{1}}x^2

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(25+x^2)y^'-2xy=0

(25 + x^{2}) y^{^{'}} - 2 x y = 0

y = 25 e^{c_{1}} + e^{c_{1}} x^{2}

Solution steps

(25 + x^{2}) y^{'} (x) - 2 x y = 0

Solve separable ODE:

y = 25 e^{c_{1}} + e^{c_{1}} x^{2}

y = 25 e^{c_{1}} + e^{c_{1}} x^{2}

\frac{d y}{d x}, x y = 54

\frac{d}{d x} (x 3 x^{2})

\int_{0}^{N} e^{(- s t)} cos (3 t) d t

\frac{\partial}{\partial x} (x sin (x^{2} y))

f^{^{'}} (x) = x^{2} - x, f (3) = \frac{19}{2}

What is the solution for (25+x^2)y^'-2xy=0 ?
The solution for (25+x^2)y^'-2xy=0 is y=25e^{c_{1}}+e^{c_{1}}x^2