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Popular >

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

Solution

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

Solution

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

Solution steps

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

Solve linear ODE:

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

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

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Frequently Asked Questions (FAQ)

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