# [SOLVED]Please check this particular solution excercise

#### ognik

##### Active member
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
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
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
Thanks Mark, found my error.

What does W|A mean please?
It is an abbreviation for Wolfram Alpha, a free online CAS that I use to check my results:

W|A y''+3y'-4y=sin(a*t) 