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How do I solve the differential equation $$7x(x-y)dy = 2(x^2+6xy-5y^2)dx$$

Is it homogeneous? I have tried taking the variables from the LHS and applying them to the RHS, making $\frac{dy}{dx}$ subject and ending up with:

$$\frac{dy}{dx} = \frac{2(x^2+6xy-5y^2)}{7x(x-y)}$$

After simplifying the numerator:

$$\frac{dy}{dx} = \frac{2[(x-y)(x+5y)+2xy]}{7x(x-y)}$$ $$\frac{dy}{dx} = \frac{2(x+5y)}{7x} + \frac{4y}{7(x-y)}$$

I have no idea how to proceed from here. Is the methodology correct so far? Any input would be appreciated.

  • 0
    Yes, it is homogeneous. Now, substitute in $u=y/x$. – 2017-01-12
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    How do I simplify the $\frac{4y}{7(x-y)}$ in terms of $\frac{dy}{dx} = F(\frac{y}{x})$ to substitute the $u=y/x$? – 2017-01-12

2 Answers 2

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Your differential equation is indeed homogeneous.

Therefore, you can substitute $y=vx$ and $\frac{dy}{dx}=x\frac{dv}{dx}+v$, where (NB: $v$ is a function of $x$). The substitution for $\frac{dy}{dx}$ is evaluated using implicit differentiation.

$$\frac{dy}{dx}=\frac{2(x^2+6xy-5y^2)}{7x(x-y)}$$

$$x\frac{dv}{dx}+v=\frac{2(x^2+6x^2v-5v^2 x^2)}{7x(x-vx)}$$

$$x\frac{dv}{dx}+v=\frac{2(1+6v-5v^2)}{7(1-v)}$$

$$x\frac{dv}{dx}=\frac{2(1+6v-5v^2)}{7(1-v)}-v$$

$$x\frac{dv}{dx}=\frac{2(1+6v-5v^2)}{7(1-v)}-\frac{7v(1-v)}{7(1-v)}$$

$$x\frac{dv}{dx}=\frac{2(1+6v-5v^2)-7v(1-v)}{7(1-v)}$$

$$x\frac{dv}{dx}=\frac{2+12v-10v^2-7v+7v^2}{7(1-v)}$$

$$x\frac{dv}{dx}=\frac{2+5v-3v^2}{7(1-v)}$$

$$x\frac{dv}{dx}=\frac{3v^2-5v-2}{7(v-1)} \tag{1}$$

You will easily notice that equation $(1)$ is separable.

Now, you can rearrange equation $(1)$, and then integrate both sides.

$$\int \frac{7(v-1)}{3v^2-5v-2}~dv=\int \frac{1}{x}~dx$$

All you have to do now, is integrate both sides, and substitute back for $v=\frac{y}{x}$.

$$\ln|2-v|+\frac{4}{3} \ln|3v+1|=\ln|x|+C$$

$$\ln\left|2-\frac{y}{x}\right|+\frac{4}{3} \ln\left|\frac{3y}{x}+1\right|=\ln|x|+C \tag{2}$$

Thus, an implicit solution to your differential equation is on equation $(2)$. I am unsure whether an explicit solution for $y(x)$ exists, however if we check on Wolfram Alpha, we get a rather strange set of explicit solutions.

  • 0
    Excuse my ignorance but how did you work out $∫\frac{7(v−1)}{3v^2−5v−2}dv$ ? I have been following the steps from Wolfram Alpha but the integration seems rather exhausting to work out, considering that this is a 12 mark examination problem and between 12 to 15 minutes should be allocated for it. Do you now of a simpler method to integrate this? Aside from that, thank you for everything, you've been a great help. – 2017-01-13
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    It is indeed quite long. I used partial fractions, however there may be a faster method which I missed. By the way, no problem. I'm happy to help. – 2017-01-13
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Starting with your equation $$\frac{dy}{dx} = \frac{2(x+5y)}{7x} + \frac{4y}{7(x-y)}$$ multiply the top and bottom of each fraction on the right, giving $$\frac{dy}{dx} = \frac{2(x+5y)}{7x}\frac{1/x}{1/x} + \frac{4y}{7(x-y)}\frac{1/x}{1/x}$$ $$\frac{dy}{dx} = \frac{2(1+5y/x)}{7} + \frac{4y/x}{7(1-y/x)}$$ Since $y'=u'x+u$ you get $$u'x+u = \frac{2(1+5u)}{7} + \frac{4u}{7(1-u)}$$ which is a separable differential equation.