# Thread: homogenous differential equation help

1. ## homogenous differential equation help

Hi

I don't understand how solve the homogenous equation

I set z=y/x therefore dy/dx=(dz/dx)x + z

however i wasn't able to fully solve it

i have attached two questions

thanks

2. $\displaystyle \frac {\mathrm{d}y}{\mathrm{d}x} = \frac {x^{2}-3y^{2}}{xy} = \frac {x}{y} - 3 \frac {y}{x}$

As you said, make the substiution $\displaystyle z = \frac {y}{x}$

then $\displaystyle \frac { \mathrm{d} y}{\mathrm{d}x} = x \frac {\mathrm{d}z}{\mathrm{d}x} + z$

so $\displaystyle x \frac {\mathrm{d}z}{\mathrm{d}x} + z = \frac {1}{z} - 3z = \frac {1-3z^{2}}{z}$

$\displaystyle \frac {z}{1-4z^{2}} \ \mathrm{d}z = \frac {1}{x} \ \mathrm{d}x$

Integrate both sides and then substitute for z.

3. Originally Posted by rpatel Hi

I don't understand how solve the homogenous equation

I set z=y/x therefore dy/dx=(dz/dx)x + z

however i wasn't able to fully solve it

i have attached two questions

thanks
For the second.
dy/dx = 1/x[y + (x)/sin(y/x)]
= y/x + [1/sin(y/x)]
let v = y/x, y = vx, dy/dx = (x)dv/dx + v
(x)dv/dx + v = v + 1/sin(v)
(x)dv/dx = 1/sin(v)
sin(v)dv = 1/x dx
-cos(v) = ln|x| + C1
cos(v) = -ln|x| + C1
v = cos^-1(C1-ln|x|)
y = xcos^-1(C1-ln|x|)

If you don't believe me check this out.
x*y*y' - y(y + x/sin(y/x)) = 0 - Wolfram|Alpha

Also, here is a link that shows you how to show that a differential equation is homogeneous Homogeneous Differential Equations: Example 2

4. Hello, rpatel!

Solve the homogenous equation: .$\displaystyle xy\frac{dy}{dx} \:=\:x^2 - 3y^2$
Divide by $\displaystyle xy\!:\;\;\frac{dy}{dx} \:=\:\frac{x}{y} - 3\frac{y}{x}$

Let: $\displaystyle z = \frac{y}{x} \quad\Rightarrow\quad y =xy \quad\Rightarrow\quad \frac{dy}{dx} \:=\:x\frac{dz}{dx} + z$

Substitute: .$\displaystyle x\frac{dz}{dx} + z \:=\:\frac{1}{z} - 3z \:=\:\frac{1-3z^2}{z}$

. . Then: .$\displaystyle \frac{dz}{dx} \:=\:\frac{1-3z^2}{z} - z \:=\:\frac{1-4z^2}{z}$

Separate variables: .$\displaystyle \frac{z}{1-4z^2}\,dz \:=\:\frac{dx}{x}$

Integrate: .$\displaystyle \int\frac{z\,dz}{1-4z^2} \;=\;\int\frac{dx}{x} \quad\Rightarrow\quad -\tfrac{1}{8}\ln\left|1-4z^2\right| \;=\;\ln x + c_1$

Multiply by -8: .$\displaystyle \ln|1-4z^2| \:=\:\text{-}8\ln x + c_2 \;=\;\ln(x^{\text{-}8}) + \ln(c_3) \;=\;\ln(c_3x^{\text{-}8})$

Exponentiate: .$\displaystyle 1 - 4z^2 \;=\;c_3x^{\text{-}8} \quad\Rightarrow\quad 1 - 4\frac{y^2}{x^2} \;=\;c_3x^{\text{-}8}$

Multiply by $\displaystyle x^2\!:\;\;x^2 - 4y^2 \;=\;c_3x^{-6} \quad\Rightarrow\quad \text{-}4y^2 \;=\;\text{-}x^2 + c_3x^{-6}$

Divide by -4: .$\displaystyle y^2 \;=\;\tfrac{1}{4}x^2 + C x^{-6}$

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