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Differential Equations (particular solutions)

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
I quote a question from Yahoo! Answers

Differential Equations help: Given Y'' + 16Y = f(x)?
1. If f(x) = 2x^2*e^(3x), give the form of Yp.
2. If f(x) = cos(2x) ,give the form of Yp.
3. If f(x) = 5x*cos(3x), give the form of Yp
I have given a link to the topic there so the OP can see my response.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
According to a well known theorem if $$y^{(n)}+a_{n-1}y^{(n-1)}+\ldots+a_1y'+a_0y=f(x)\;(E)$$ with $f(x)=e^{\alpha x}\left(P_k(x)\cos \beta x+Q_r(x)\sin \beta x\right)$, a particular solution of $(E)$ has the form: $$y_p(x)=x^se^{\alpha x}\left(\tilde{P_d}(x)\cos \beta x+\tilde{Q_d}(x)\sin \beta x\right)$$ where $(i)$ $\tilde{P_d},\tilde{Q_d}$ are polynomials of degree $d=\max \left\{{k,r}\right\}$. $(ii)$ $s$ is the order of $\alpha +\beta i$ as a root of the characteristic equation $\lambda^n+a_{n-1}\lambda^{n-1}+\ldots+a_1\lambda+a_0=0$. In our case, $\lambda^2+16=0\Leftrightarrow \alpha+\beta i=\pm4i$. So,

1. If $f(x)=2x^2e^{3x}$, then $y_p(x)=e^{3x}(ax^2+bx+c)$.

2. If $f(x)=\cos 2x$, then $y_p(x)=a\cos 2x+b\sin 2x$.

3. If $f(x)=5x\cos 3x$, then $y_p(x)=(ax+b)\cos \color{red}3x+(cx+d)\sin \color{red}3x$.
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
One may also employ the annihilator method to determine the forms of the particular solutions.

1.) \(\displaystyle f(x)=2x^2e^{3x}\)

If we observe that the differential operator $A$ defined as:

\(\displaystyle A\equiv(D-3)^3\)

annihilates $f(x)$, then we know the particular solution, with the characteristic root $r=3$ of multiplicity 3, must have the form:

\(\displaystyle y_p(x)=c_1e^{3x}+c_2xe^{3x}+c_3x^2e^{3x}=\left(c_1+c_2x+c_3x^2 \right)e^{3x}\)

2.) \(\displaystyle f(x)=\cos(2x)\)

If we observe that the differential operator $A$ defined as:

\(\displaystyle A\equiv D^2+4\)

annihilates $f(x)$, then we know the particular solution, with the characteristic roots $r=\pm2i$, must have the form:

\(\displaystyle y_p(x)=c_1\cos(2x)+c_2\sin(2x)\)

3.) \(\displaystyle f(x)=5x\cos(3x)\)

If we observe that the differential operator $A$ defined as:

\(\displaystyle A\equiv(D^2+9)^2\)

annihilates $f(x)$, then we know the particular solution, with the characteristic roots $r=\pm3i$, both of multiplicity 2, must have the form:

\(\displaystyle y_p(x)=\left(c_1+c_2x \right)\cos(3x)+\left(c_3+c_4x \right)\sin(3x)\)

In all 3 cases, we see that the characteristic roots of the differential operator are different from those of the associated homogeneous equation.