Linear homogeneous D.E. with constant coefficients - known solutions

1. Aug 12, 2014

etf

My task is to find Linear homogeneous D.E. with constant coefficients which has solutions:
$$\\\varphi 1(x)=x^2,\varphi 2(x)=e^{-3x},\varphi 3(x)=cos(5x)$$ Any idea?

Last edited: Aug 12, 2014
2. Aug 12, 2014

etf

I think that here we have root $$\\\lambda 1=0$$ with multiplicity 3 (from $$\\\varphi 1(x)$$), root $$\\\lambda 2=-3$$ with multiplicity 1 (from $$\\\varphi 2(x)$$).

3. Aug 12, 2014

Zondrina

I assume you mean second order.

The solution $x^2$ should come from a Cauchy Euler equation I believe. So I think you need to find an equation of that form by multiplying the roots together to "work backwards" to the indicial equation.

4. Aug 12, 2014

slider142

It is possible to have $\varphi_1(x) = x^2$ as a solution to a linear homogeneous equation with constant coefficients. In addition, a linear homogeneous differential equation has as many linearly independent solutions as the order of the equation. Since the 3 functions you listed are linearly independent, the equation we are looking for must be at least of order 3.
As a hint, note that any linear combination of the solutions must also be a solution. Thus, the 3-parameter family of functions $\varphi(x) = C_1x^2 + C_2e^{-3x} + C_3\cos(5x)$ is a solution to the unknown differential equation. As a further hint, try taking this solution's derivatives. Can you find a combination using constant coefficients that is homogeneous?

5. Aug 12, 2014

pasmith

You're missing $\cos(5x) = (e^{5ix} + e^{-5ix})/2$.

Thus the auxiliary equation must be a polynomial with roots 0 (with multiplicity 3), -3, 5i and -5i.

6. Aug 12, 2014

Zondrina

Wow, I totally forgot equations higher than second order existed ^_^.

7. Aug 12, 2014

etf

I calculated characteristic polynomial as $$\\(\lambda -\lambda1)^{3}(\lambda-\lambda2)(\lambda-\lambda3)(\lambda-\lambda4)=(\lambda-0)^{3}(\lambda+3)(\lambda-(0-5i))(\lambda-(0+5i))=...$$
$$=\lambda^{6}+3\lambda^{5}+25\lambda^{4}+75\lambda^{3}$$ so D.E. is $$\\y^{(6)}+3y^{(5)}+25y^{(4)}+75y^{(3)}$$

Last edited by a moderator: Aug 13, 2014
8. Aug 12, 2014

ehild

It looks correct. The result is the same using Slider142's hint.

ehild

9. Aug 12, 2014

LCKurtz

Except that that isn't an equation.

Last edited by a moderator: Aug 13, 2014
10. Aug 12, 2014

ehild

Oh, =0 is missing! I did not notice.

ehild

11. Aug 13, 2014

etf

Forgot to write = 0

12. Aug 13, 2014

etf

Here is another one (this one is not with constan coefficients):
Find linear homogeneous second order D.E. which has general solution $$\\\varphi (x)=C1\frac{sinx}{\sqrt{x}}+C2\frac{\cos{x}}{\sqrt{x}}, x>0$$
Any idea?

13. Aug 13, 2014

ehild

What is the differential equation for y(x)=√(x) φ(x)?

ehild

14. Aug 13, 2014

etf

I found it. $$\\-\frac{1}{x}y''-\frac{1}{x^{2}}y'+(\frac{1}{4x^{3}}-\frac{1}{x})y=0$$

15. Aug 13, 2014

ehild

Congratulation!
Have you used my hint? It is very easy with that.

ehild

16. Aug 13, 2014

etf

To be honest, I didn't understand your hint :(
I solved it using Wronskian:

$$\\y1(x)=\frac{\sin{x}}{\sqrt{x}},$$
$$\\y2(x)=\frac{\cos{x}}{\sqrt{x}},$$
$$\\W(y1,y2)=\begin{vmatrix} y1(x) & y2(x) & y \\ y1'(x) & y2'(x) & y'(x) \\ y1''(x)& y2''(x) & y''(x) \end{vmatrix}=0$$

However, this is a little bit complicated.

17. Aug 13, 2014

ehild

$$\varphi (x)=C1\frac{sinx}{\sqrt{x}}+C2\frac{\cos{x}}{\sqrt{x}}$$

Multiply by √x:

$$\sqrt{x}\varphi (x)=C1\sin(x)+C2\cos(x)$$

Define a new function $y(x)=\sqrt{x}\varphi (x)$. The general solution of a linear homogeneous differential equation for y(x) is $y(x)= C1\sin(x)+C2\cos(x)$. But that is the solution of the well known differential equation $y''+ y=0$.

Find the second derivative of $y(x)=\sqrt{x}\varphi (x)$ in terms of x and φ and substitute into the equation $y''+ y=0$...

ehild

18. Aug 13, 2014

etf

Wow
Your method is definitely much better :)
Thanks!