Consider 2nd order differential equation:(adsbygoogle = window.adsbygoogle || []).push({});

[itex]

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

[/itex]

Let b and c be the roots of the quadratic [itex]a_{2}x^2 + a_{1}x + a_{0}[/itex]

Verify that when b = c then [itex]xe^{bx}[/itex] is a solution of the Differential equation.

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Ok so I differentiate y a couple times and get

[itex]

y = xe^{bx}

[/itex]

[itex]

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

[/itex]

[itex]

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

[/itex]

Substituting into the equation yields:

[itex]

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

[/itex]

Now if I rearrange the quadratic equation for a0

[itex]

a_{0} = -a_{2}b^2 - a_{1}b

[/itex]

then substituting that in I get

[itex]

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

[/itex]

[itex]

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

[/itex]

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

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# Homework Help: Verify solutions of differential equation

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