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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 : $(x/4)[x\sin x-\cos x]$

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  • $\begingroup$ Did you try to use the method? On another subject: since there are no constants in the answer, you should be given initial values. $\endgroup$ – Dennis Gulko Nov 23 '12 at 10:10
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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 $C_2(x)$.

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  • $\begingroup$ 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)? $\endgroup$ – Garett Nov 23 '12 at 11:27
  • $\begingroup$ 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 - $y_p$! $\endgroup$ – Dennis Gulko Nov 23 '12 at 11:31
  • $\begingroup$ @DennisGulko is this the variation of parameters method ? $\endgroup$ – theenigma017 May 28 at 21:42
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Beside to Dennis's approach, you can use the annihilator's method as well http://www.utdallas.edu/dept/abp/PDF_Files/DE_Folder/Annihilator_Method.pdf. 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 $y_P(x)$.

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  • $\begingroup$ $\ddot\smile\quad +1 \quad $ $\endgroup$ – Namaste Apr 7 '13 at 0:13
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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 $$ \begin{align} y &=e^{ix}\int e^{-ix}\left(e^{-ix}\left(\frac{x^2}4+\frac{ix}2+c_1\right)-e^{ix}\left(\frac{ix}4+\frac18\right)\right)\,\mathrm{d}x\\ &=e^{-ix}\left(\frac{ix^2}8-\frac x8\right)-e^{ix}\left(\frac{ix^2}8+\frac x8\right)+c_1^\prime e^{-ix}+c_2^\prime e^{ix}\\ &=\frac{x^2}4\sin(x)-\frac x4\cos(x)+c_1^{\prime\prime}\cos(x)+c_2^{\prime\prime}\sin(x)\tag{10} \end{align} $$

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