Solving Homogeneous ODE: $\displaystyle x(y-3x)\frac{dy}{dx}=2y^2-9xy+8x^2$

In summary, a homogeneous differential equation is one in which all the terms can be expressed in the same degree of the independent variable. To solve such an equation, a substitution method can be used. An ODE is considered homogeneous if it can be written in a specific form. The general solution to a homogeneous ODE involves an arbitrary function, but initial conditions can be used to obtain a unique solution.
  • #1
Guest2
193
0
I'm trying to solve $\displaystyle x(y-3x)\frac{dy}{dx} = 2y^2-9xy+8x^2$

Let $y = vx$ then $\displaystyle \frac{dy}{dx}= v+x\frac{dv}{dx}$ and I end up with

$\displaystyle \log(cx) = \frac{1}{2}\log(y^2/x^2-6y/x+8)$

Is this correct and what am I supposed to do after this?
 
Last edited:
Physics news on Phys.org
  • #2
Re: Differential equation

It looks good so far! I would multiply both sides by 2, exponentiate, and then solve for $y$. Don't forget to check by plugging back into the original.
 

FAQ: Solving Homogeneous ODE: $\displaystyle x(y-3x)\frac{dy}{dx}=2y^2-9xy+8x^2$

What is a homogeneous differential equation?

A homogeneous differential equation is one in which all the terms can be expressed in the same degree of the independent variable. This means that all the terms in the equation have the same power of the independent variable, and there are no constants or terms with different powers.

How do I solve a homogeneous differential equation?

To solve a homogeneous differential equation, you can use a substitution method where you let y = vx, and then solve for v. You can then use this value of v to find the solution for y.

How do I know if an ODE is homogeneous?

An ODE is homogeneous if it can be written in the form of $\displaystyle F(x,y) = F(tx,ty)$, where t is a constant. In other words, if all the terms in the equation can be factored out as a power of the independent variable, then the ODE is homogeneous.

What is the general solution to a homogeneous ODE?

The general solution to a homogeneous ODE is given by $\displaystyle y = x^k\Phi\left(\frac{y}{x}\right)$, where k is the degree of the ODE and $\Phi$ is an arbitrary function. This solution can be obtained by using the substitution method mentioned earlier.

Can I use initial conditions to find a particular solution to a homogeneous ODE?

Yes, you can use initial conditions to find a particular solution to a homogeneous ODE. However, since the general solution contains an arbitrary function, you will need to use the initial conditions to determine the value of this function and obtain a unique solution.

Similar threads

MHB -2.2.35 Show that dy/dx=(x+3y)/(x-y) is homogeneous. and....
Replies
2
Views
1K
MHB -b.2.2.1 separate variables y'=\dfrac{x^2}{y}
Replies
4
Views
2K
MHB -2.2.12 IVP y'=2x/(y+x^2y), y(0)=-2
Replies
2
Views
2K
MHB Solve DE for y: $\displaystyle y^\prime +y = xe^{-x}+1$
Replies
2
Views
1K
MHB -2.2.20 IVP interval....trig subst y^2(1-x^2)^{1/2} \,dy=\arcsin{x}\,dx
Replies
4
Views
2K
MHB 2.2.8 de xy'+2y=sin x,y(pi)=1/pi find solution
Replies
2
Views
2K
MHB Solve $\displaystyle 3y'-2y=e^{-\pi/2}\quad y(0)=a$: Find $a_0$ Exactly
Replies
1
Views
2K
MHB 2.2.5 de equation with y(1)=1/2
Replies
4
Views
1K
MHB -aux.2.2.4 de y'+(\cot x)y&=2 \csc x; y(\pi/2)=a
Replies
2
Views
1K
MHB Soln of IVP: $y = \frac{e^x + 1 - e}{x}, \space x>0$
Replies
1
Views
1K

Hot Threads

Recent Insights

Back
Top