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

[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?

 

[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 $$\\\varphi 2(x)$$).

 

[STEMucator]

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.

 

[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?

 

[pasmith]

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 $$\\\varphi 2(x)$$).

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.

 

[STEMucator]

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

My bad.

 

[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 $$\\y^{(6)}+3y^{(5)}+25y^{(4)}+75y^{(3)}$$

 

[ehild]

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

ehild

 

[LCKurtz]

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 $$\\y^{(6)}+3y^{(5)}+25y^{(4)}+75y^{(3)}$$

Except that that isn't an equation.

 

[ehild]

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

ehild

 

[etf]

Except that that isn't an equation.

Forgot to write = 0

 

[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?

 

[ehild]

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

ehild

 

[etf]

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

 

[ehild]

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

ehild

 

[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.

 

[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

 

[etf]

Wow
Your method is definitely much better :)
Thanks!