Verify solutions of differential equation

[coldturkey]

Consider 2nd order differential equation:

$a_{2}y''(x) + a_{1}y'(x) + a_{0}y(x) = 0$

Let b and c be the roots of the quadratic $a_{2}x^2 + a_{1}x + a_{0}$

Verify that when b = c then $xe^{bx}$ is a solution of the Differential equation.
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Ok so I differentiate y a couple times and get

$y = xe^{bx}$

$y' = xbe^{bx} + e^{bx}$

$y'' = xb^2e^{bx}+2be^{bx}$

Substituting into the equation yields:

$a_{2}(xb^2e^{bx}+2be^{bx}) + a_{1}(xbe^{bx} + e^{bx}) + a_{0}(xe^{bx}) = 0$

Now if I rearrange the quadratic equation for a0
$a_{0} = -a_{2}b^2 - a_{1}b$

then substituting that in I get

$a_{2}(xb^2e^{bx}+2be^{bx}) + a_{1}(xbe^{bx} + e^{bx}) + (-a_{2}b^2 - a_{1}b)(xe^{bx}) = 0$

$= a_{2}2be^{bx} + a_{1}e^{bx} = 0$

This is where I get stuck, there must be something I've missed which I can use to further simplify. I did the proof for $y = e^{bx}$ in the same way and it worked out easy. Can anybody give me a hint on what to do next? thanks

 

[nrqed]

Consider 2nd order differential equation:

$a_{2}y'' + a_{1}y' + a_{0}y = 0$

Let b and c be the roots of the quadratic $a_{2}x^2 + a_{1}x + a_{0}$

Verify that when b = c then $xe^{bx}$ is a solution of the Differential equation.
-----------------------------

Doesn't that imply that $a_0 = {a_1^2 \over 4 a_2}$? It does not seem to be what you used.

Also, from th eway the question is posed, the b in the exponent has nothing to do with the b in "the roots are b and c", right?

 

[coldturkey]

They are the same b and c as I take it.

Where do you get $a_0 = {a_1^2 \over 4 a_2}$ from. I just assumed that it was trying to say $a_{2}b^2 + a_{1}b + a_{0} = 0$

 

[nrqed]

They are the same b and c as I take it.

Where do you get $a_0 = {a_1^2 \over 4 a_2}$ from. I just assumed that it was trying to say $a_{2}b^2 + a_{1}b + a_{0} = 0$

That's not the way I intrepreted the sentence, but I may be wrong.

Consider $a_2 x^2 + a_1 x + a_0$. The two roots are $x=-{a_1 \over 2} \pm {1 \over 2} {\sqrt{ a_1^2 - 4 a_0 a_2}}$.
Let's call these two roots b and c. If the two roots are equal, then the term under the square root must be zero.

That's the only way I can make sense of "the two roots of the above quadratic are equal".

But I may be completely off. Sorry.

 

[nrqed]

That's not the way I intrepreted the sentence, but I may be wrong.

Consider $a_2 x^2 + a_1 x + a_0$. The two roots are $x=-{a_1 \over 2} \pm {1 \over 2} {\sqrt{ a_1^2 - 4 a_0 a_2}}$.
Let's call these two roots b and c. If the two roots are equal, then the term under the square root must be zero.

That's the only way I can make sense of "the two roots of the above quadratic are equal".

But I may be completely off. Sorry.

I forgot to divide by a_2 in writing the two roots.

I think this is what they meant to say , it seems to be working (if in addition I assume that the b equal to the unique root is the b appearing in the exponent)

 

[nrqed]

I forgot to divide by a_2 in writing the two roots.

I think this is what they meant to say , it seems to be working (if in addition I assume that the b equal to the unique root is the b appearing in the exponent)

assuming that the sqaure root vanishes, then the single root is -a_1 / 2 a_2. setting this equal to b, one gets a_1 = -2 b a_2.


In addition, imposing a_0 = a_1^2/(4 a_2) gives a_0 = b^2 a_2.

Using those two expressions for a_0 and a_1 in terms of a_2, you will get that everything cancels out.

Patrick

 

[coldturkey]

ahh i can see what you did now. I'll have a crack at it, thanks for your help