Author Topic: TT1 Problem 1 (noon)  (Read 1800 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2563
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
TT1 Problem 1 (noon)
« on: October 16, 2018, 05:30:35 AM »
Find integrating factor and then a general solution of ODE
\begin{equation*}
y\bigl(4\ln(x)+4\ln(y)+1 \bigr) + x\bigl(\ln(x)+\ln(y)+1\bigr) y'=0
\end{equation*}
 
Also, find a solution satisfying $y(1)=1$.

Shengying Yang

  • Jr. Member
  • **
  • Posts: 10
  • Karma: 17
    • View Profile
Re: TT1 Problem 1 (noon)
« Reply #1 on: October 16, 2018, 07:22:50 AM »
$$M=y(4lnx + 4lny +1)=y(4ln(xy)+1),N=x(lnx+lny+1)=x(ln(xy)+1)$$
$$M_y =4ln(xy) +5, N_x=ln(xy)+2$$
Since $M_y\ne N_x​$​, it is not exact.
$$R=\frac{M_y-N_x}{N}=\frac{3ln(xy)+3}{x(ln(xy)+1)} = \frac{3(ln(xy)+1)}{x(ln(xy)+1)}=\frac{3}{x}$$
R is a function of x only, so there is an integrating factor of the form
$$\mu=e^{\int R dx}=e^{\int \frac{3}{x} dx}=e^{3lnx }=x^3$$
multiply the differential equation by $\mu​$
$$x^3y(4ln(xy) +1)+x^4(ln(xy)+1)y'=0$$
Since ​$M_y = N_x​$, It is exact now. There exist a ​ $\psi(x,y)​$ such that ​$\psi_x=M, \psi_y=N​$ .
$$\psi=\int M dx=\int x^3y(4ln(xy) +1)dx=x^4yln(xy)+h(y)$$
$$\psi_y=x^4ln(xy)+x^4+h'(y)=x^4(ln(xy)+1)=N$$
$$∴h'(y)=0, h(y)=C$$
$$Therefore, \psi(x,y)= x^4yln(xy)=C$$
as x=1, y=1, we get C=0
$$∴x^4yln(xy)=0$$

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2563
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: TT1 Problem 1 (noon)
« Reply #2 on: October 16, 2018, 07:29:44 AM »
Can you simplify the answer. Also ln , sin, ... should be escaped: \ln x

Shengying Yang

  • Jr. Member
  • **
  • Posts: 10
  • Karma: 17
    • View Profile
Re: TT1 Problem 1 (noon)
« Reply #3 on: October 16, 2018, 07:53:51 AM »
Here is my simplified answer.
$$M_y =4\ln (xy) +5 \ne N_x=\ln (xy)+2$$
∴not exact.
$$R=\frac{M_y-N_x}{N}= \frac{3(\ln (xy)+1)}{x(\ln (xy)+1)}=\frac{3}{x}$$
$$\mu=e^{\int \frac{3}{x} dx}=x^3$$
$$∴x^3y(4\ln (xy) +1)+x^4(\ln (xy)+1)y'=0$$
There exist a ​ $\psi(x,y)​$ such that ​$\psi_x=M, \psi_y=N​$ .
$$\psi=\int x^3y(4\ln (xy) +1)dx=x^4y\ln (xy)+h(y)$$
$$\psi_y=x^4\ln (xy)+x^4+h'(y)$$
$$∴h'(y)=0, h(y)=C$$
$$∴\psi(x,y)= x^4y\ln (xy)=C$$
as x=1, y=1, we get C=0
$$∴x^4y\ln (xy)=0$$

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 2563
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
Re: TT1 Problem 1 (noon)
« Reply #4 on: October 18, 2018, 04:02:19 AM »
Shengying, I asked to simplify the answer, not the solution in general: $x^4y\ln (xy)=0\implies \ln (xy)=0$ due to condition $y(1)=1$, then $xy=1\implies y=x^{-1}$.