Original Problem: Solve Differential Equation

$$\left(x^2 \sec{\left(y+1\right)} \tan (y+1)-3xe^{y+1}+\frac{2yx^{2}}{y^2+1}\right) \frac{dy}{dx}+3x \sec(y+1)-6e^{y+1}+3x \ln{\left(y^2+1\right)}=0$$

I am working on a differential equation and got to this pint when doing the integrating factor:

$$\frac { x \sec(y+1) \tan(y+1)-3e^{y+1}+\frac{4xy}{y^2+1} } { x^2 \sec(y+1) \tan(y+1)-3x e^{y+1}+\frac{2yx^{2}}{y^2+1} }$$

Can I go any further. I know that if I multiply the top by x all the variables are the same on top and bottom. Then I divide long division and get 2. So is this the answer to this integrating factor?

  • $\begingroup$ Can you add the original problem? $\endgroup$ – Amzoti Sep 30 '13 at 15:33
  • $\begingroup$ Just put original problem up! $\endgroup$ – user95801 Sep 30 '13 at 16:07
  • $\begingroup$ I formatted your equations to make them readable (you can check your post to see how I did it, for the future). Please double check if they are correctly displayed. $\endgroup$ – Lisa Sep 30 '13 at 16:38
  • $\begingroup$ Just went through them only one error but I fixed it thank you! $\endgroup$ – user95801 Sep 30 '13 at 16:41
  • $\begingroup$ Do you realize that your statement 'if I multiply the top by x all the variables are the same on top and bottom' means that your fraction equals precisely $1/x$?! $\endgroup$ – automaton 3 Sep 30 '13 at 22:25

So we have $Pdx+Qdy=0$, where $$ P=3x \sec(y+1)-6e^{y+1}+3x \ln{\left(y^2+1\right)} $$ $$ Q=x^2 \sec{\left(y+1\right)} \tan (y+1)-3xe^{y+1}+\frac{2yx^{2}}{y^2+1} $$ It is not an exact equation, but may be we can obtain an exact equation multiplying by a suitable factor:

$$ P_y=3x \tan(y+1) \sec(y+1)-6e^{y+1}+3x \frac{2y}{y^2+1} $$ $$ Q_x=2x \tan(y+1) \sec(y+1)-3e^{y+1}+\frac{4yx}{y^2+1} $$ Note that $$ P_y-Q_x=x\tan(y+1)\sec(y+1)-3e^{y+1}+ \frac{2yx}{y^2+1}=\frac{Q}{x} $$ Then $$ \frac{\alpha'(x)}{\alpha(x)}=\frac{P_y-Q_x}{Q}=\frac{1}{x} $$ $$ \frac{d\alpha}{\alpha}=\frac{dx}x $$

Therefore the factor $\alpha=x$ becomes the equation in an exact equation. It remains to integrate the new equation...

We need to find the potential $g$ of the vector field $(x\,P(x,y),x\,Q(x,y))$:

$$ g(x,y)=\int 3x^2 \sec(y+1)-6x\,e^{y+1}+3x^2 \ln{\left(y^2+1\right)} \,dx +c(y) = \\ =x^3 \sec(y+1) -3x^2\,e^{y+1}+x^3\ln{\left(y^2+1\right)} +c(y) = \\ =x^3 \sec(y+1) -3x^2\,e^{y+1}+x^3\ln{\left(y^2+1\right)} +c $$

Finally, the solutions are given by the equation: $$ x^3 \sec(y+1) -3x^2\,e^{y+1}+x^3\ln{\left(y^2+1\right)} +c =0 $$

  • $\begingroup$ Doesnt Py−Qx =Q/x not xQ, and the last equation i have never seen before it it like the Py-Qx/Q equation? $\endgroup$ – user95801 Oct 1 '13 at 14:30
  • $\begingroup$ Fixed. Thanks for the comment. $\endgroup$ – Pocho la pantera Oct 1 '13 at 15:48
  • $\begingroup$ Thanks for all the help!!! $\endgroup$ – user95801 Oct 2 '13 at 14:36

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