# Homogeneous Differential Equation

#### nasi112

##### New member
$$\displaystyle x^2 \frac{d^2y}{{dx}^2} + x \frac{dy}{dx} + y = 0$$

This differential equation can be solved easily when we make $$\displaystyle x = e^t$$

But I cannot solve it when the coefficients are not 1!

Something like this

$$\displaystyle 2x^2 \frac{d^2y}{{dx}^2} + 3x \frac{dy}{dx} + 4y = 0$$

Is there a way to solve it? Any ideas?

#### markraz

##### Full Member
can you use Ax=B to solve? null space problem?

#### Jomo

##### Elite Member
Please show us how you would solve the first problem you stated. We will go from there.

#### Romsek

##### Senior Member
$$\displaystyle x^2 \dfrac{d^2y}{dx^2} + x \dfrac{dy}{dx} + y = 0$$

Now let
$$\displaystyle u=\log(x)\\ \dfrac{d^2y}{du^2} = \dfrac{f''(u)-f'(u)}{x^2} \\ \dfrac{dy}{du} = \dfrac{f'(u)}{x}$$

Substituting back into the original equation we get

$$\displaystyle y''(u) - y'(u) + y'(u) + y(u) = y''(u) + y(u) = 0$$

This is now trivial to solve in $$\displaystyle u$$ so solve it and the solution to the original equation will be $$\displaystyle y(\log(x))$$

#### HallsofIvy

##### Elite Member
$$\displaystyle x^2 \frac{d^2y}{{dx}^2} + x \frac{dy}{dx} + y = 0$$

This differential equation can be solved easily when we make $$\displaystyle x = e^t$$
No, it can't! You may be confusing it with the solution $$\displaystyle y= Ce^{mx}$$ for equations with constant coefficients.

But I cannot solve it when the coefficients are not 1!

Something like this

$$\displaystyle 2x^2 \frac{d^2y}{{dx}^2} + 3x \frac{dy}{dx} + 4y = 0$$

Is there a way to solve it? Any ideas?
Equivalently to doing the u= log(x) substitution, as Romsek shows, try $$\displaystyle y= x^s$$. Then $$\displaystyle y'= sx^{s-1}$$ and $$\displaystyle y''= s(s-1)x^{s-2}$$. The equation becomes $$\displaystyle 2s(s-1)x^s+ 3sx^s+ 4x^s= x^s(2s(s- 1)+ 3s+ 4)= 0$$. In order that this be 0 for all x, we must have $$\displaystyle 2s(s-1)+ 3s+ 4= 2s^2+ s+ 4= 0.$$

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#### nasi112

##### New member
I am sorry for the late reply.

Thank you so much guys for giving me some ideas, but I have said that I could solve the first problem when $$\displaystyle x = e^t$$

the solution for $$\displaystyle x^2\frac{d^2y}{{dx}^2} + x\frac{dy}{dx} + y = 0$$ is

$$\displaystyle y(x) = c_1 \cos(\ln(x)) + c_2 \sin(\ln(x))$$

the difficulty arises when the differential equation looks like this

$$\displaystyle 2x^2\frac{d^2y}{{dx}^2} + 3x\frac{dy}{dx} + 4y = 0$$

$$\displaystyle x = e^t$$ does not work here

can you use Ax=B to solve? null space problem?
You're asking me? I have no idea what you mean!

Please show us how you would solve the first problem you stated. We will go from there.
It is not difficult to show the solution for the first problem.

$$\displaystyle x^2 \dfrac{d^2y}{dx^2} + x \dfrac{dy}{dx} + y = 0$$

Now let
$$\displaystyle u=\log(x)\\ \dfrac{d^2y}{du^2} = \dfrac{f''(u)-f'(u)}{x^2} \\ \dfrac{dy}{du} = \dfrac{f'(u)}{x}$$

Substituting back into the original equation we get

$$\displaystyle y''(u) - y'(u) + y'(u) + y(u) = y''(u) + y(u) = 0$$

This is now trivial to solve in $$\displaystyle u$$ so solve it and the solution to the original equation will be $$\displaystyle y(\log(x))$$
I know how to solve the first problem. I am asking about the second problem.

No, it can't! You may be confusing it with the solution $$\displaystyle y= Ce^{mx}$$ for equations with constant coefficients.

Equivalently to doing the u= log(x) substitution, as Romsek shows, try $$\displaystyle y= x^s$$. Then $$\displaystyle y'= sx^{s-1}$$ and $$\displaystyle y''= s(s-1)x^{s-2}$$. The equation becomes $$\displaystyle 2s(s-1)x^s+ 3sx^s+ 4x^s= x^s(2s(s- 1)+ 3s+ 4)= 0$$. In order that this be 0 for all x, we must have $$\displaystyle 2s(s-1)+ 3s+ 4= 2s^2+ s+ 4= 0.$$
Yes I could, and I have already done it.

#### Romsek

##### Senior Member
well the answer seems to be

$$\displaystyle y(x) = \frac{c_1 \sin \left(\frac{1}{4} \sqrt{31} \log (x)\right)}{\sqrt[4]{x}}+\frac{c_2 \cos \left(\frac{1}{4} \sqrt{31} \log (x)\right)}{\sqrt[4]{x}}$$

I confess I don't know how that might be derived.

#### nasi112

##### New member
Thanks a lot Romsek.