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Popular >

y^'=-(2y)/(x+1)+2x

Solution

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

Solution

y = \frac{x ^{4}}{x ^{2} + 2 x + 1} + \frac{2 x ^{3}}{x ^{2} + 2 x + 1} + \frac{x ^{2}}{x ^{2} + 2 x + 1} - \frac{x ^{4}}{2 ( x ^{2} + 2 x + 1 )} - \frac{2 x ^{3}}{3 ( x ^{2} + 2 x + 1 )} + \frac{c _{1}}{x ^{2} + 2 x + 1}

Solution steps

y^{'} (x) = - \frac{2 y}{x + 1} + 2 x

Solve linear ODE:

y = \frac{x ^{4}}{x ^{2} + 2 x + 1} + \frac{2 x ^{3}}{x ^{2} + 2 x + 1} + \frac{x ^{2}}{x ^{2} + 2 x + 1} - \frac{x ^{4}}{2 ( x ^{2} + 2 x + 1 )} - \frac{2 x ^{3}}{3 ( x ^{2} + 2 x + 1 )} + \frac{c _{1}}{x ^{2} + 2 x + 1}

y = \frac{x ^{4}}{x ^{2} + 2 x + 1} + \frac{2 x ^{3}}{x ^{2} + 2 x + 1} + \frac{x ^{2}}{x ^{2} + 2 x + 1} - \frac{x ^{4}}{2 ( x ^{2} + 2 x + 1 )} - \frac{2 x ^{3}}{3 ( x ^{2} + 2 x + 1 )} + \frac{c _{1}}{x ^{2} + 2 x + 1}

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Frequently Asked Questions (FAQ)

What is the solution for y^'=-(2y)/(x+1)+2x ?
The solution for y^'=-(2y)/(x+1)+2x is