What is the solution for (1+t^2)y^'+4ty=t ?

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

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(1+t^2)y^'+4ty=t

(1 + t^{2}) y^{^{'}} + 4 t y = t

y = \frac{c _{1}}{4 ( t ^{4} + 2 t ^{2} + 1 )} + \frac{1}{4}

Solution steps

(1 + t^{2}) y^{'} (t) + 4 t y = t

Solve separable ODE:

y = \frac{c _{1}}{4 ( t ^{4} + 2 t ^{2} + 1 )} + \frac{1}{4}

y = \frac{c _{1}}{4 ( t ^{4} + 2 t ^{2} + 1 )} + \frac{1}{4}

x \to - 4 lim (\frac{1}{( x + 4 ) ^{4}})

\frac{d}{d x} (10 x + 4 y)

8 tan (2 x)

\frac{d}{d x} (x^{2} - 64)

x \to 28 lim (\frac{x ^{2}}{4} - \frac{7}{x})

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