   My Coding > Mathematics > First order ordinary differential equations > Solving xy' - 2y - x^2 = 0

# Solving xy' - 2y - x^2 = 0

In this lecture, I will show you how to solve the $$xy\prime - 2y - x^2 = 0$$ equation. This linear differential equation should be solver via the integration factor method and now I will show ho to do it.

The first step is to convert this equation to a canonical form.

$y\prime - \frac{2}{x}y = x$

And now we need to introduce an integrational factor:

$Iy\prime - I\frac{2}{x}y = Ix$

Product rule:

$(Iy)\prime = Iy\prime + I\prime y$

Due to the arbitrary form, of $$I$$, we can require the equality of the terms in the sum, but it is important to take care about the sign of these terms:

$I\prime = -I\frac{2}{x}$

and solve it against $$I$$:

$\frac{\mathop{dI}}{I} = -\frac{2\mathop{dx}}{x}$

$\int \frac{\mathop{dI}}{I} = -\int \frac{2\mathop{dx}}{x} \Rightarrow \ln{\left|I\right|} = -2\ln{\left|x\right|} + C$

$\ln{\left|I\right|} = \ln{Ax^{-2)}$

$e^{\ln{\left|I\right|}} = e^{\ln{Ax^{-2}}}$

$\boxed{I = x^{-2}}$

From our equation we can write:

$(Iy)\prime = Ix = x^{-2}x = frac{1}{x}$

Now integrate both parts and:

$Iy = \int \frac{\mathop{dx}}{x} + C$

$x^{-2}y = \ln{\left|x\right|} + C$

and finally:

$\boxed{y = x^2\ln{\left|x\right|} + Cx^2}$

For more details, how to solve $$x^2y\prime + 2xy = \cos^2{x}$$, please watch this video:

Published: 2023-05-14 05:58:53
Updated: 2023-05-14 06:01:46

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