the question is:

$$(y^2+xy^3)\mathrm dx + (5y^2-xy+y^3\sin(y))\mathrm dy = 0$$

can any body tell me how to solve this linear equation?? when I tried to solve this the expression of integrating factor becomes too much difficult, may be i calculated it wrong... Any help will be appreciated. Thanks!

  • $\begingroup$ Linear equation? The unknown function must be $x=x(y)$, then. $\endgroup$ – Giuseppe Negro Apr 20 '11 at 19:43
  • $\begingroup$ @Sadia: Looks to me to be an exact equation. $\endgroup$ – night owl Jul 4 '11 at 12:05

First I'd divide through by $y^2$ to make your life easier. Then the partial differential equation for the integrating factor becomes

$$xu+\frac{\partial u}{\partial y}(1+xy)=-\frac{1}{y}u+\frac{\partial u}{\partial x}(t-\frac{x}{y}+y\sin y)\;.$$

That happens to have a solution with $\frac{\partial u}{\partial x}=0$, so you can determine an integrating factor $u(y)$ from it.

  • $\begingroup$ Partial differential equation? OP has a linear DE (in $x$), there's a formula for the integrating factor, no need to solve a PDE, is there? $\endgroup$ – Gerry Myerson Apr 26 '11 at 10:01
  • $\begingroup$ @Gerry: This is not what I'd call a linear DE (in $x$); it contains $x\mathrm dx$. If you mean the formula given e.g. here: en.wikipedia.org/wiki/…, I don't think it applies in this case; at least I don't see how to bring this equation into that form (after swapping $x$ and $y$). $\endgroup$ – joriki Apr 26 '11 at 10:19
  • $\begingroup$ sorry, must have posted while my brain was in neutral. $\endgroup$ – Gerry Myerson Apr 26 '11 at 12:59

$(y^2+xy^3)~dx+(5y^2-xy+y^3\sin y)~dy=0$

$(xy^3+y^2)~dx=(xy-5y^2-y^3\sin y)~dy$

$\left(x+\dfrac{1}{y}\right)\dfrac{dx}{dy}=\dfrac{x}{y^2}-\dfrac{5}{y}-\sin y$

Let $u=x+\dfrac{1}{y}$ ,

Then $x=u-\dfrac{1}{y}$


$\therefore u\left(\dfrac{du}{dy}+\dfrac{1}{y^2}\right)=\dfrac{1}{y^2}\left(u-\dfrac{1}{y}\right)-\dfrac{5}{y}-\sin y$

$u\dfrac{du}{dy}+\dfrac{u}{y^2}=\dfrac{u}{y^2}-\dfrac{1}{y^3}-\dfrac{5}{y}-\sin y$

$u\dfrac{du}{dy}=-\dfrac{1}{y^3}-\dfrac{5}{y}-\sin y$

$u~du=\left(-\dfrac{1}{y^3}-\dfrac{5}{y}-\sin y\right)~dy$

$\int u~du=\int\left(-\dfrac{1}{y^3}-\dfrac{5}{y}-\sin y\right)~dy$

$\dfrac{u^2}{2}=\dfrac{1}{2y^2}-5\ln y+\cos y+c$

$\left(x+\dfrac{1}{y}\right)^2=\dfrac{1}{y^2}-10\ln y+2\cos y+C$


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