My Math Forum differential equation use substitution methode

 Differential Equations Ordinary and Partial Differential Equations Math Forum

 June 3rd, 2012, 01:52 AM #1 Member   Joined: Apr 2012 Posts: 47 Thanks: 0 differential equation use substitution methode for this equation i use substitution methode. but my answer is not right. dont know where i make a mistake, please someone can help me to check. thanks! right answer is y=4x ln(cx)
 June 3rd, 2012, 02:43 AM #2 Member   Joined: Apr 2012 Posts: 47 Thanks: 0 Re: differential equation use substitution methode i have an other question $y'=1-\frac{y}{x}$ $u=\frac{y}{x}$ $y=1-u$ $y=u.x$ $y'=u#39;.x+u$ last step i dont get it, where is u' and +u get from? what is the u'?
 June 3rd, 2012, 02:50 AM #3 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: differential equation use substitution methode U=Y*1/x+4 x*U-4=Y Its the mistake in your previous question
 June 3rd, 2012, 03:05 AM #4 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: differential equation use substitution methode x*y'=y+4*x x*dy=y*dx+4x*dx x*dy-y*dx=4x*dx (x*dy-y*dx)/(x^2) = (4/x)*dx d(y/x) = (4/x)*dx, integrating both sides, y/x+c1 = 4ln(x)+c2 y=4xln(x)+Cx where C=c1-c2 My C = your 4ln(c)
June 3rd, 2012, 03:49 AM   #5
Member

Joined: Apr 2012

Posts: 47
Thanks: 0

Re: differential equation use substitution methode

Quote:
 Originally Posted by mathbalarka U=Y*1/x+4 x*U-4=Y Its the mistake in your previous question
thanks alot mathbalarka for helping me out. do you know this question?

Quote:
 Originally Posted by zell^ i have an other question $y'=1-\frac{y}{x}$ $u=\frac{y}{x}$ $y=1-u$ $y=u.x$ $y'=u#39;.x+u$ last step i dont get it, where is u' and +u get from? what is the u'?

 June 3rd, 2012, 04:03 AM #6 Math Team   Joined: Nov 2010 From: Greece, Thessaloniki Posts: 1,990 Thanks: 133 Math Focus: pre pre pre pre pre pre pre pre pre pre pre pre calculus Re: differential equation use substitution methode [color=#000000]What mathbalarka has done..... remember that $\left(\frac{f(x)}{g(x)}\right)'=\frac{f'(x )g(x)-f(x)g#39;(x)}{g(x)^{2}}$, now for $x\neq 0$ $xy'(x)-y(x)=4x\Rightarrow \frac{xy'(x)-y(x)}{x^2}=\frac{4}{x}\Rightarrow \left(\frac{y(x)}{x}\right)'=\frac{4}{x}\Right arrow\int\left(\frac{y(x)}{x}\right)#39;\;dx=\int \frac{4}{x}\;dx\Rightarrow \frac{y(x)}{x}=4\ln|x|+c\Rightarrow$ $y(x)=4x\ln|x|+cx$.[/color]
 June 3rd, 2012, 04:16 AM #7 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: differential equation use substitution methode y'=1-y/x x*y'=x-y x*y'+y=x x*y'+x'*y=x (xy)'=x d(xy)=x*dx integrating, xy+c1=x^2/2+c2 xy=x^2/2+C y=x/2+C/x
 June 3rd, 2012, 04:21 AM #8 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory Re: differential equation use substitution methode If you want, I can tell you the solution of $xY'(x) = ax+bY(x)$ a,b are constants Note: sorry for writing answers of previous questions without using $\LaTeX$
 June 3rd, 2012, 08:42 AM #9 Member   Joined: Apr 2012 Posts: 47 Thanks: 0 Re: differential equation use substitution methode thx problems solved. now got an other one problem. how do i found out C is just C not lnC? how can i recognize it after intergration? example: $\int \frac{1}{y}dy=\int \frac{1}{x}dx$ $ln y= ln x+ ln C$ other one $\int \frac{1}{y}=\int \frac{1}{x^{2}}dx$ $ln y=-\frac{1}{x}+ln C$ confused
June 3rd, 2012, 10:02 AM   #10
Math Team

Joined: Mar 2012
From: India, West Bengal

Posts: 3,871
Thanks: 86

Math Focus: Number Theory
Re: differential equation use substitution methode

Quote:
 Originally Posted by zell^ thx problems solved. now got an other one problem. how do i found out C is just C not lnC? how can i recognize it after intergration? example: $\int \frac{1}{y}dy=\int \frac{1}{x}dx$ $ln y= ln x+ ln C$ other one $\int \frac{1}{y}=\int \frac{1}{x^{2}}dx$ $ln y=-\frac{1}{x}+ln C$ confused
Can you explain your question more simply? I cant understand
According to my calculations,
The genarel solution of $xY'(x) = Ax + BY(x)$ is:
$Y(x) = e^{(B-2)C} \frac{x^B}{B(B-1)} - \frac{A}{B-1} x
\text{Please forgive me if I am wrong}$

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post sydfremmer Differential Equations 6 February 12th, 2013 12:45 PM alexmath Differential Equations 2 November 21st, 2012 11:49 AM jrkcolorado Linear Algebra 1 September 22nd, 2012 12:37 PM zell^ Differential Equations 3 June 8th, 2012 02:42 PM germanaries Real Analysis 0 March 18th, 2008 12:28 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top