Solving a 2nd Order ODE: P^2 - 4xP +6y = 0

[Saladsamurai]

Homework Statement

$$\left(\frac{dy}{dx}\right)^2 - 4x\frac{dy}{dx} + 6y = 0$$

Homework Equations

A common approach we have used for similar problems has been to let P = dy/dx

The Attempt at a Solution

Doing so we have:

$$P^2 - 4xP +6y = 0$$

$$\Rightarrow 6y = 4P(x - P)$$

Differentiating gives:

$$6P = 4\left[P(1 - \frac{dP}{dx}) +\frac{dP}{dx}(x - P)\right] = 0$$Now usually we try to factor this and solve each factor as a linear 1st order EQ in P. However, I am having trouble seeing a nice way to factor this, that makes each factor linear. All I can get to is

$$-2\left[P+2\frac{dP}{dx} - 2x\frac{dP}{dx} + 2P\frac{dP}{dx}\right] = 0$$Any thoughts on what to do with the bracketed term to get 2 linear EQs out of the deal?

Thanks

 

[lippyka]

A:$$6P = 4\left[P(1 - \frac{dP}{dx}) +\frac{dP}{dx}(x - P)\right] = 0$$$$\Rightarrow 6P=4\left[P-\frac{dP}{dx}+x\frac{dP}{dx}-P\frac{dP}{dx}\right]$$$$\Rightarrow 6P=4\left[P+x\frac{dP}{dx}-\left(1+P\right)\frac{dP}{dx}\right]$$$$\Rightarrow \frac{6P}{4}=P+x\frac{dP}{dx}-\left(1+P\right)\frac{dP}{dx}$$$$\Rightarrow P+x\frac{dP}{dx}-\left(1+P\right)\frac{dP}{dx}=0$$Divide by $\frac{dP}{dx}$$$\Rightarrow P+x-\left(1+P\right)=0$$$$\Rightarrow P+x=1+P$$$$\Rightarrow P=1+x$$$$\Rightarrow \frac{dP}{dx}=1$$