Find the general solution of the differential equation:



$$x^2+\int 2xydx=\int 4dx\tag2$$



The answer from the book is $$y=\frac{4}{x}+\frac{c}{x^2}$$

  • $\begingroup$ Where does the line $x^2+\int 2xy\,\mathrm{d}x=\int 4\,\mathrm{d}x$ come from? $\endgroup$ – user10354138 Jun 22 at 9:28
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    $\begingroup$ The passage after $\frac{dy}{dx}x^2+2xy=4$ looks like nonsense $\endgroup$ – Saucy O'Path Jun 22 at 9:29
  • $\begingroup$ I am trying to separate the variable but it is not working $\endgroup$ – user546244 Jun 22 at 9:31
  • $\begingroup$ Seems like you were trying to integrate both sides, rather. $\endgroup$ – Saucy O'Path Jun 22 at 9:32
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    $\begingroup$ With (1) write $(x^2y)'=4$. $\endgroup$ – Nosrati Jun 22 at 9:33

$$\frac{dy}{dx}+\frac{2y}{x}=\frac{4}{x^2}\qquad . . . . . . (1)$$

This is a first order linear differential equation.

Integrating factor (I.F.) is $$e^{\int \frac{2}{x} dx}=e^{2\log x}=x^2$$ So multiplying both side of $(1)$ by I.F. $$x^2\frac{dy}{dx}+2xy=4\implies \frac{d}{dx}(x^2y)=4$$ and then integrating$$x^2y=4x+c\implies y=\frac{4}{x}+\frac{c}{x^2}\qquad \text{where $c$ is integrating constant.}$$


You can solve firstly the homogeneous differential equation.



$\int \frac{dy}{y}=-\int\frac{2}{x}dx$

$\ln(y)=-2\cdot \ln(x)+c$

$\ln(y)=- \ln(x^2)+c$


$y_H=C\cdot \frac{1}{x^2}$

Variation of the constant.

$y_I=C(x)\cdot \frac{1}{x^2}$

Differentiating (product rule)

$y_I^{'}=C^{'}(x)\cdot \frac{1}{x^2}-C^{}(x)\cdot \frac{2}{x^3}$

Plugging into the differential equation

$C^{'}(x)\cdot \frac{1}{x^2}\underbrace{-C^{}(x)\cdot \frac{2}{x^3}+2\cdot C(x)\cdot \frac{1}{x^3}}_{=0}=\frac{4}{x^2}$

$C^{'}(x)\cdot \frac{1}{x^2}=\frac{4}{x^2}$

$C^{'}(x)=4\Rightarrow C(x)=4x$

Thus the solution is $y=y_H+y_I=C\cdot \frac{1}{x^2}+4x\cdot \frac{1}{x^2}$

$$y=\frac{C}{x^2}+ \frac{4}{x}$$


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