Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

By using the method of undetermined coeficiens, find the particular solution $y_p$ for this inhomogenous differential equation. $$y''+y= \sin x + x\cos x$$

I have find the roots which are $\pm i$.

and the complementary function $y_c= (C_1 \cos x + C_2\sin x)$

The answer given is : $(x/4)[x\sin x-\cos x]$

share|cite|improve this question
Did you try to use the method? On another subject: since there are no constants in the answer, you should be given initial values. – Dennis Gulko Nov 23 '12 at 10:10
up vote 1 down vote accepted

By setting $C_1=C_1(x)$ and $C_2=C_2(x)$, you have $y=C_1(x)\cos x+C_2(x)\sin x$. Substitute this into the equation. You will have a system of equations on the derivatives of $C_1,C_2$. You should assume that the sum of all terms containing $C_1''(x)$ and $C_2''(x)$ is zero. Now find $C_1(x)$ and $C_2(x)$. I.e: $$\begin{array}{l} y=C_1(x)\cos x+C_2(x)\sin x\\ y'=-C_1(x)\sin x+C_2(x)\cos x+C_1'(x)\cos x+C_2'(x)\sin x\\ \begin{align*}y''=&-C_1(x)\cos x-C_2(x)\sin x-C_1'(x)\sin x+C_2'(x)\cos x\\&-C_1'(x)\sin x+C_2'(x)\cos x+C_1''(x)\cos x+C_2''(x)\sin x\end{align*}\\ \end{array}$$ $$\begin{align*} \sin x+x\cos x &=y''+y=C_1(x)\cos x+C_2(x)\sin x-C_1(x)\cos x-C_2(x)\sin x\\&-C_1'(x)\sin x+C_2'(x)\cos x-C_1'(x)\sin x+C_2'(x)\cos x+C_1''(x)\cos x+C_2''(x)\sin x \end{align*}$$ $$\left\{\begin{array}{l} 2C_2'(x)\cos x-2C_1'(x)\sin x=\sin x+x\cos x\\ C_1''(x)\cos x+C_2''(x)\sin x=0 \end{array}\right.$$ $$C_2'(x)=\frac12\left(\tan x+x+C_1'(x)\tan x\right)$$ $$C_2''(x)=\frac12\left(\frac{1}{1+x^2}+1+C_1'(x)\frac{1}{1+x^2}+C_1''(x)\tan x\right)$$ $$0=C_1''(x)\cos x+\frac12\left(\frac{\sin x}{1+x^2}+\sin x+C_1'(x)\frac{\sin x}{1+x^2}+C_1''(x)\cos x\right)$$ $$C_1''(x)+C_1'(x)\frac13\frac{\tan x}{1+x^2}+\frac13\frac{\tan x}{1+x^2}+\frac13\tan x=0$$ Now find $C_1(x)$ and $C_2(x)$.

share|cite|improve this answer
Hi Dennis.Actually i want to know particular solution y(p) not complematary function y(c). So, the method you propose actually seems not right because you comparing the coefficient for y(c) NOT y(p). So how am I suppose to know particular solution y(p)? – Garett Nov 23 '12 at 11:27
That is the way to find the particular solution. When you will find $C_1$ and $C_2$, substitute them back to $y_c$ and you will get the particular solution - $y_p$! – Dennis Gulko Nov 23 '12 at 11:31

Beside to Dennis's approach, you can use the annihilator's method as well According to this way you will have: $$(D^2+1)^3y=0$$ so we can guess $$y(x)=y_p(x)+y_c(x)\\=A\cos(x)+B\sin(x)+Ex\cos(x)+Fx\sin(x)+Gx^2\cos(x)+Hx^2\sin(x)$$ Now pick up the $y_c(c)$ from the solution . You will have $y_P(x)$.

share|cite|improve this answer
$\ddot\smile\quad +1 \quad $ – amWhy Apr 7 '13 at 0:13

Consider the differential equation $$ \left(\frac{\mathrm{d}}{\mathrm{d}x}+a\right)y=f(x)\tag{1} $$ This can be solved using an integrating factor. Suppose $$ g'(x)=ag(x)\tag{2} $$ then $$ \frac{\mathrm{d}}{\mathrm{d}x}(g(x)y)=g(x)y'+ag(x)y=g(x)f(x)\tag{3} $$ Equation $(3)$ is simply $(1)$ multiplied by $g(x)$. Note that $(2)$ is the same as $$ \frac{\mathrm{d}}{\mathrm{d}x}\log(g(x))=a\tag{4} $$ which is satisfied by $$ g(x)=e^{ax}\tag{5} $$ plugging $(5)$ back into $(3)$ yields $$ \frac{\mathrm{d}}{\mathrm{d}x}\left(e^{ax}y\right)=e^{ax}f(x)\tag{6} $$ which becomes $$ y=e^{-ax}\int e^{ax}f(x)\,\mathrm{d}x\tag{7} $$

$y''+y=\sin(x)+x\cos(x)$ is simply $$ \left(\frac{\mathrm{d}}{\mathrm{d}x}+i\right)\left(\frac{\mathrm{d}}{\mathrm{d}x}-i\right)y=\sin(x)+x\cos(x)\tag{8} $$ Using $(7)$ once with $a=i$ gives $$ \begin{align} \left(\frac{\mathrm{d}}{\mathrm{d}x}-i\right)y &=e^{-ix}\int e^{ix}(\sin(x)+x\cos(x))\,\mathrm{d}x\\ &=e^{-ix}\left(\frac{x^2}4+\frac{ix}2+c_1\right)-e^{ix}\left(\frac{ix}4+\frac18\right)\tag{9} \end{align} $$ Using $(7)$ again with $a=-i$ gives $$ \begin{align} y &=e^{ix}\int e^{-ix}\left(e^{-ix}\left(\frac{x^2}4+\frac{ix}2+c_1\right)-e^{ix}\left(\frac{ix}4+\frac18\right)\right)\,\mathrm{d}x\\ &=e^{-ix}\left(\frac{ix^2}8-\frac x8\right)-e^{ix}\left(\frac{ix^2}8+\frac x8\right)+c_1^\prime e^{-ix}+c_2^\prime e^{ix}\\ &=\frac{x^2}4\sin(x)-\frac x4\cos(x)+c_1^{\prime\prime}\cos(x)+c_2^{\prime\prime}\sin(x)\tag{10} \end{align} $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.