Eliminating constants in a second derivative of a function

[bergausstein]

Eliminate the constants

$y=c_1 e^{ax}cosbx +c_2e^{ax}sinbx$

since there's two constants I took the derivative twice

$y'=c_1\left ( -e^{ax}b\sin{bx} + ae^{ax}\cos{bx}\right )+ c_2\left ( e^{ax}b\cos{bx} +ae^{ax}\sin{bx}\right )$

$y''=c_1\left [ \left ( -e^{ax}b^2\cos{bx}-ae^{ax}b\sin{bx}\right) + \left (-ae^{ax}b\sin{bx}+a^2e^{ax}\cos{bx} \right ) \right ]+c_2\left [ \left ( -e^{ax}b^2\sin{bx}+ae^{ax}b\cos{bx}\right )+\left ( ae^{ax}b\cos{bx}+a^2e^{ax}\sin{bx}\right ) \right ]$now I don't know what to do next. please help me!

 

[MarkFL]

Let's factor and combine like terms and line them up as follows:

\(\displaystyle y=c_1e^{ax}\cos(bx)+c_2e^{ax}\sin(bx)\)

\(\displaystyle y'=c_1e^{ax}\left(a\cos(bx)-b\sin(bx) \right)+c_2e^{ax}\left(b\cos(bx)+a\sin(bx) \right)\)

\(\displaystyle y''=c_1e^{ax}\left(\left(a^2-b^2 \right)\cos(bx)-2ab\sin(bx) \right)+c_2e^{ax}\left(2ab\cos(bx)+\left(a^2-b^2 \right)\sin(bx) \right)\)

Now, can you find two factors such that when you multiply the first and second equations by these factors, the resulting sum of the three equations is zero?

 

[HallsofIvy]

Eliminate the constants

$y=c_1 e^{ax}cosbx +c_2e^{ax}sinbx$

since there's two constants I took the derivative twice

$y'=c_1\left ( -e^{ax}b\sin{bx} + ae^{ax}\cos{bx}\right )+ c_2\left ( e^{ax}b\cos{bx} +ae^{ax}\sin{bx}\right )$

$y''=c_1\left [ \left ( -e^{ax}b^2\cos{bx}-ae^{ax}b\sin{bx}\right) + \left (-ae^{ax}b\sin{bx}+a^2e^{ax}\cos{bx} \right ) \right ]+c_2\left [ \left ( -e^{ax}b^2\sin{bx}+ae^{ax}b\cos{bx}\right )+\left ( ae^{ax}b\cos{bx}+a^2e^{ax}\sin{bx}\right ) \right ]$now I don't know what to do next. please help me!

Multiply y' by 2a, multiply y by $$a^2+ b^2$$ and add:
y''+ 2ay'+ (a^2+ b^2)y= 0

- - - Updated - - -

Eliminate the constants

$y=c_1 e^{ax}cosbx +c_2e^{ax}sinbx$

since there's two constants I took the derivative twice

$y'=c_1\left ( -e^{ax}b\sin{bx} + ae^{ax}\cos{bx}\right )+ c_2\left ( e^{ax}b\cos{bx} +ae^{ax}\sin{bx}\right )$

$y''=c_1\left [ \left ( -e^{ax}b^2\cos{bx}-ae^{ax}b\sin{bx}\right) + \left (-ae^{ax}b\sin{bx}+a^2e^{ax}\cos{bx} \right ) \right ]+c_2\left [ \left ( -e^{ax}b^2\sin{bx}+ae^{ax}b\cos{bx}\right )+\left ( ae^{ax}b\cos{bx}+a^2e^{ax}\sin{bx}\right ) \right ]$now I don't know what to do next. please help me!

Multiply y' by 2a, multiply y by $$a^2+ b^2$$ and add:
$$y''+ 2ay'+ (a^2+ b^2)y$$

 

[bergausstein]

Let's factor and combine like terms and line them up as follows:

\(\displaystyle y=c_1e^{ax}\cos(bx)+c_2e^{ax}\sin(bx)\)

\(\displaystyle y'=c_1e^{ax}\left(a\cos(bx)-b\sin(bx) \right)+c_2e^{ax}\left(b\cos(bx)+a\sin(bx) \right)\)

\(\displaystyle y''=c_1e^{ax}\left(\left(a^2-b^2 \right)\cos(bx)-2ab\sin(bx) \right)+c_2e^{ax}\left(2ab\cos(bx)+\left(a^2-b^2 \right)\sin(bx) \right)\)

Now, can you find two factors such that when you multiply the first and second equations by these factors, the resulting sum of the three equations is zero?

I can't find it. :(

 

[MarkFL]

I can't find it. :(

You need not search, you have been provided with the factors. (Tmi)