# Thread: What is homogenous differential equation? how can we apply solve this homogenous diff

1. ## What is homogenous differential equation? how can we apply solve this homogenous diff

What is homogenous differential equation? solve this homogenous differential equation: (x^2+y^2)dx-2xy dy=0 and i want to understand the basic how can i apply homogenious differential equation term on this math because i am new with this homogenous term.Thanks in advance.

2. ## Re: What is homogenous differential equation? how can we apply solve this homogenous

Where did you get this problem? Are you taking a differential equations course or reading a differential equation text book? In either case, you should have seen a definition of "homogeneous differential equation". Actually, there are two different concepts of "homogenous" differential equations, one apply only to first order equations, such as this, the other to linear differential equations (in the case of a "first order, linear, differential equations" they are the same). A first order differential equation, of the form M(x,y)dy+ N(x,y)dy= 0, is "homogenous" if and only if $\displaystyle \frac{M(\lambda x, \lambda y)}{N(\lambda x, \lambda y)}= \frac{M}{N}$ for any number $\displaystyle \lambda$.

In this equation, $\displaystyle M= x^2+ y^2$ and $\displaystyle N= -2xy$ so that $\displaystyle \frac{M(\lambda x, \lambda y)}{N(\lambda x, \lambda y)}$$= \frac{(\lambda x)^2+ (\lambda y)^2}{-2(\lambda x)(\lambda y)}=$$ \frac{\lambda^2(x^2+ y^2)}{-2\lambda^2xy}= \frac{x^2+ y^2}{-xy}=$$\frac{M(x, y)}{N(x,y)}. The point of such "homogeneous" differential equations is that we can write them in terms of the single variable \displaystyle v= \frac{y}{x}. In this case, dividing each part of the equation by \displaystyle x^2 we have \displaystyle \frac{x^2+ y^2}{x^2}dy- \frac{2xy}{x^2}dx=$$\displaystyle \left(1+ \left(\frac{y}{x}\right)\right)^2)dx- 2\frac{y}{x}dx$. If we let $\displaystyle v= \frac{y}{x}$ then $\displaystyle xv= y$ so, differentiating both sides using the product rule, vdx+ xdv= 2y, so we can write the equation as $\displaystyle (1+ v^2)dx- 2v(vdx+ xdv)= (1- 2v+ v^2)dx- 2xvdv= 0$ so that $\displaystyle \frac{dx}{x}= \frac{2v}{v^2- 2v+ 1}dv$. Integrate that.

3. ## Re: What is homogenous differential equation? how can we apply solve this homogenous

How can i check this equation is this equation is homogenuos 1st order differential equation or not and how can i apply 1st order homogenious differential equation on this math:
(x^2+y^2)dx-2xy dy=0 and how to check homogeneous 1st order equation term on the math.please explain elaborately.

4. ## Re: What is homogenous differential equation? how can we apply solve this homogenous

You could start by reading my response to your first post where I told you exactly what to do and what the answer is! If there are parts you do not understand, tell us what those parts are. What do you know about "mathematics", "Calculus", and "differential equations"?

I asked before "Are you taking a differential equations course or reading a differential equation text book?" Could you answer that question so we have a better idea what kind of responses will help you.

5. ## Re: What is homogenous differential equation? how can we apply solve this homogenous Originally Posted by HallsofIvy Where did you get this problem? Are you taking a differential equations course or reading a differential equation text book? In either case, you should have seen a definition of "homogeneous differential equation". Actually, there are two different concepts of "homogenous" differential equations, one apply only to first order equations, such as this, the other to linear differential equations (in the case of a "first order, linear, differential equations" they are the same). A first order differential equation, of the form M(x,y)dy+ N(x,y)dy= 0, is "homogenous" if and only if $\displaystyle \frac{M(\lambda x, \lambda y)}{N(\lambda x, \lambda y)}= \frac{M}{N}$ for any number $\displaystyle \lambda$.

In this equation, $\displaystyle M= x^2+ y^2$ and $\displaystyle N= -2xy$ so that $\displaystyle \frac{M(\lambda x, \lambda y)}{N(\lambda x, \lambda y)}$$= \frac{(\lambda x)^2+ (\lambda y)^2}{-2(\lambda x)(\lambda y)}=$$ \frac{\lambda^2(x^2+ y^2)}{-2\lambda^2xy}= \frac{x^2+ y^2}{-xy}=$$\frac{M(x, y)}{N(x,y)}. The point of such "homogeneous" differential equations is that we can write them in terms of the single variable \displaystyle v= \frac{y}{x}. In this case, dividing each part of the equation by \displaystyle x^2 we have \displaystyle \frac{x^2+ y^2}{x^2}dy- \frac{2xy}{x^2}dx=$$\displaystyle \left(1+ \left(\frac{y}{x}\right)\right)^2)dx- 2\frac{y}{x}dx$. If we let $\displaystyle v= \frac{y}{x}$ then $\displaystyle xv= y$ so, differentiating both sides using the product rule, vdx+ xdv= 2y, so we can write the equation as $\displaystyle (1+ v^2)dx- 2v(vdx+ xdv)= (1- 2v+ v^2)dx- 2xvdv= 0$ so that $\displaystyle \frac{dx}{x}= \frac{2v}{v^2- 2v+ 1}dv$. Integrate that.
Hello,
How did you calculate -2xvdv? What is equal to dv? How did you calculate $(1+v^2)dx-2v(vdx+xdv)?$Please explain.

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