No idea to do this differential equation queation.

[vkash]

Homework Statement

x²y₁+xyy₁-6y²=0
find solution for this differential equation.

y₁ mean dy/dx

How to do this question. I have no idea.

 

[Stephen Tashi]

x2y1+xyy1-6y2=0

$$x^2 \frac{dy}{dx} + xy\frac{dy}{dx} - 6y^2 = 0$$

I don't know how to do it systematically, but one may reason about it.

If we assume $y$ is an nth degree polynomial in x, the degrees of the 3 terms are n+1, 2n, 2n. For the terms to cancel, we need n+1 = 2n, so n = 1.

So try letting $y = Cx + D$ and solve for $C$ and $D$.

If there is a "proper" way of doing it, I hope someone tells us.

 

[obafgkmrns]

Hint: Try differentiating x^2*y^2.

 

[Ray Vickson]

$$x^2 \frac{dy}{dx} + xy\frac{dy}{dx} - 6y^2 = 0$$

I don't know how to do it systematically, but one may reason about it.

If we assume $y$ is an nth degree polynomial in x, the degrees of the 3 terms are n+1, 2n, 2n. For the terms to cancel, we need n+1 = 2n, so n = 1.

So try letting $y = Cx + D$ and solve for $C$ and $D$.

If there is a "proper" way of doing it, I hope someone tells us.

Maple 14 gets the solution as
y = (x/5)*[1+Z(z)^5], where Z(x) is a solution of the equation 5z^6 - cxz^5 - cx=0, and c is an arbitrary constant. When you take c = 0 the solution is pretty simple; otherwise, it is likely horrible.

RGV

 

[Stephen Tashi]

Maple 14 gets the solution as
y = (x/5)*[1+Z(z)^5], where Z(x) is a solution of the equation 5z^6 - cxz^5 - cx=0, and c is an arbitrary constant.

I wonder if human beings are supposed to do it by "integrating factors"

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

Is not an "exact" differential equation, but perhaps there is an integrating factor M(x,y) so that:
$M(x,y)(x^2 + xy)\frac{dy}{dx} - 6M(x,y)y^2 = 0$ is exact.

 

[vkash]

sorry sorry sorry this is wrong question.
The correct question.correct one is
x²y₁²+xyy₁-6y²=0

 

[SammyS]

Factor this.

Let u = x·y',

then u² + uy -6y² = 0

Factor the left hand side.

(u+3y)(u-2y)=0

Solve for u, then put x·y' back in for u .

Can you take it from there?