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[SOLVED] Please check this particular solution excercise

ognik

Active member
Feb 3, 2015
471
Given $y'' + 3y'-4y= sin \omega t $, I used an ansatz of $y_p = A sin \omega t + B cos \omega t$

$\therefore y' = A \omega cos \omega t -B \omega sin \omega t, y'' = -A \omega^2 sin \omega t - B \omega^2 cos \omega t $

Substituting and equating coefficients, I get $ -A \omega^2 - 3B \omega - 4A = 1 (a), -B \omega^2 +3A \omega -4B = 0 (b)$

From (b) $A = \frac{B \omega^2 + 4B}{3\omega} $, substituting this into (a) gives $B = \frac{3}{35 \omega - \omega^3} $

I'd appreciate if someone could check this please?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
According to W|A, the particular solution is:

\(\displaystyle -\frac{\omega^2+4}{\left(\omega^2+1\right)\left(\omega^2+16\right)}\sin(\omega t)-\frac{3\omega}{\left(\omega^2+1\right)\left(\omega^2+16\right)}\cos(\omega t)\)

You have assumed the correct form for the particular solution:

\(\displaystyle y_p(t)=A\sin(\omega t)+B\cos(\omega t)\)

And so we compute:

\(\displaystyle y_p'(t)=A\omega\cos(\omega t)-B\omega\sin(\omega t)\)

\(\displaystyle y_p''(t)=-A\omega^2\sin(\omega t)-B\omega^2\cos(\omega t)\)

Substituting for $y_p$ into the given ODE, and appropriately arranging, we obtain:

\(\displaystyle \left(-A\omega^2-3B\omega-4A\right)\sin(\omega t)+\left(-B\omega^2+3A\omega-4B\right)\cos(\omega t)=1\cdot\sin(\omega t)+0\cdot\cos(\omega t)\)

Equating coefficients, we obtain the system:

\(\displaystyle A\omega^2+3B\omega+4A=-1\)

\(\displaystyle B\omega^2-3A\omega+4B=0\)

Solving the first equation for $A$, we obtain:

\(\displaystyle A=-\frac{1+3B\omega}{\omega^2+4}\)

Substituting into the second, we obtain:

\(\displaystyle B\omega^2+3\frac{1+3B\omega}{\omega^2+4}\omega+4B=0\)

\(\displaystyle B\omega^2\left(\omega^2+4\right)+3\omega(1+3B\omega)+4B\left(\omega^2+4\right)=0\)

\(\displaystyle B\omega^4+17B\omega^2+16B=-3\omega\)

\(\displaystyle B\left(\omega^2+1\right)\left(\omega^2+16\right)=-3\omega\)

\(\displaystyle B=-\frac{3\omega}{\left(\omega^2+1\right)\left(\omega^2+16\right)}\)

And so we then obtain:

\(\displaystyle A=-\frac{1+3\left(-\dfrac{3\omega}{\left(\omega^2+1\right)\left(\omega^2+16\right)}\right)\omega}{\omega^2+4}\)

\(\displaystyle A=-\frac{\left(\omega^2+1\right)\left(\omega^2+16\right)-9\omega^2}{\left(\omega^2+4\right)\left(\omega^2+1\right)\left(\omega^2+16\right)}=-\frac{\left(\omega^2+4\right)^2}{\left(\omega^2+4\right)\left(\omega^2+1\right)\left(\omega^2+16\right)}=-\frac{\omega^2+4}{\left(\omega^2+1\right)\left(\omega^2+16\right)}\)

And so we find we agree with W|A, which is generally a good thing. :)
 

ognik

Active member
Feb 3, 2015
471
And so we find we agree with W|A, which is generally a good thing. :)
Thanks Mark, found my error.

What does W|A mean please?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775