# Find the general solution of $y'' − 2y' + 5y = e^x \cos(2x)$

I've racked my head against this for hours. Finding the complementary solution (homogenous solution) is fairly simple and I got $$y_c = e^x[ C_1\sin(2x)+ C_2\cos(2x) ].$$

But I am stuck on finding the particular solution to complete the general solution.

I tried the undetermined coefficient approach but everything would keep cancelling out and I would get a 0 on one side.

• What techniques do you know for finding particular solutions when the inhomogeneous part is of the same form as the solution to the homogeneous part? Aug 16 '19 at 3:13
• You can make the problem simpler via the substitution $z = e^x y$. The basic issue will remain, though: The inhomogeneous part of your equation is itself a solution to the homogeneous equation. Aug 16 '19 at 3:13
• @kittercatter I don't know any special technique for that. I just tried undetermined coefficient and variation of parameter. Aug 16 '19 at 3:23
• @MerajHaq I think variation of parameters should get you there, maybe need an ansatz or guess to get things correct. Could you update your question with your variation of parameter attempt? Aug 16 '19 at 3:30
• @kittercatter I tried variation of parameters. Unless I made a mistake I did not get the answer. Which is supposed to be y = (1/4)xe^xsin(2x) +yc. Aug 16 '19 at 3:38

$$y'' − 2y' + 5y = e^x \cos(2x)\implies (D^2-2D+5)y=e^x \cos(2x)\qquad \text{where}\quad D\equiv \frac{d}{dx}$$

For particular integral (P.I.),

P.I.$$~=\frac{1}{D^2-2D+5}~e^x \cos(2x)$$

$$~~~~~~~= ~e^x~\frac{1}{(D+1)^2-2(D+1)+5}~\cos(2x)$$

$$~~~~~~~= ~e^x~\frac{1}{D^2+4}~\cos(2x)$$

$$~~~~~~~=~\frac{x}{4}~e^x~\sin(2x)$$

So the general solution is $$y(x)= e^x[ C_1\sin(2x)+ C_2\cos(2x) ]~+~\frac{x}{4}~e^x~\sin(2x)\qquad \text{where}\quad C_1,~C_2~\text{are constants.}$$

Note$$~ 1:$$

For the Particular Integral (i.e., P.I.) there are some general rules

$$1.$$ $$\frac{1}{D + a} \phi (x) = e^{-ax}\int e^{ax}\phi(x)$$

$$2.$$ $$\frac{1}{f(D)} e^{ax} \phi(x) = e^{ax}\frac{1}{f(D+a)} \phi(x)$$

$$3.$$ $$\frac{1}{f(D)} x^{n} \sin ax =$$Imaginary part of $$e^{iax}\frac{1}{f(D+ia)} x^n$$

$$4.$$ $$\frac{1}{f(D)} x^{n} \cos ax =$$Real part of $$e^{iax}\frac{1}{f(D+ia)} x^n$$

$$5.$$ $$\frac{1}{f(D)} x^{n} (\cos ax + i\sin ax) = \frac{1}{f(D)} x^n e^{iax}=e^{iax}\frac{1}{f(D+ia)} x^n$$

Note$$~ 2:$$ For the Particular Integral (i.e., P.I.) of trigonometric functions you have to follow the following rules:

If $$f(D)$$ can be expressed as $$\phi(D^2)$$ and $$\phi(-a^2)\neq 0$$, then

$$1.$$ $$\frac{1}{f(D)} \sin ax=\frac{1}{\phi(D^2)} \sin ax = \frac{1}{\phi(-a^2)} \sin ax$$

$$2.$$ $$\frac{1}{f(D)} \cos ax=\frac{1}{\phi(D^2)} \cos ax = \frac{1}{\phi(-a^2)} \cos ax$$

Note: If $$f(D)$$ can be expressed as $$\phi(D^2)=D^2+a^2$$, then $$\phi(-a^2)= 0$$.

$$1.$$ $$\frac{1}{f(D)} \sin ax =\frac{1}{\phi(D^2)} \sin ax=x\frac{1}{\phi'(D^2)} \sin ax= x \frac{1}{2D} \sin ax= -\frac{x}{2a} \cos ax$$.

$$2.$$ $$\frac{1}{f(D)} \cos ax =\frac{1}{\phi(D^2)} \cos ax=x\frac{1}{\phi'(D^2)} \cos ax= x \frac{1}{2D} \cos ax= \frac{x}{2a} \sin ax$$.

where $$\phi'(D^2)\equiv\frac{d}{dD}\phi(D^2)$$

• Very very thorough Aug 16 '19 at 4:31

Use undetermined coefficients

I think your supposed to let $$y_p=Axe^{x}\cos \left(2x\right)+bxe^{x}\sin \left(2x\right)$$

then take $$y'\left(p\right)$$ and $$y"\left(p\right)$$ plug em in for $$y"\left(p\right)+2y'\left(p\right)+y\left(p\right)$$ and solve for $$A and B$$ to find particular solution

But taking all these derivatives would be an extreme hassle.

After simplifying $$y"\left(p\right)+2y'\left(p\right)+y\left(p\right)=e^x\cos \left(2x\right)$$ I end up with

$$4Be^x\cos 2x-4Ae^x\sin \left(2x\right)=e^x\cos \left(2x\right)$$

So $$4B=1$$ and $$A=0$$ then $$y_p=\frac{1}{4}xe^x\sin \left(2x\right)$$

Remember product rule for three functions is $$\left(fgh\right)'=f'gh+fg'h+fgh'$$

Also remember to multiply an $$x$$ to the guess for the form of the particular solution since the normal guess of $$y_p=Ae^{x}\cos \left(2x\right)+be^{x}\sin \left(2x\right)$$ appears in the complementary solution.

• thanks for the answer but I tried this. Unless I made a mistake everything cancels out when plugging in so it does not work. Aug 16 '19 at 3:58
• Use \sin and \cos (MathJax tutorial) Aug 16 '19 at 4:27
• @MerajHaq given that the inhomogeneous piece has the same form as the solution to the homogeneous diff-eq a natural ansatz/guess is to add $x$ to the front. en.wikipedia.org/wiki/… see that last sentence Aug 16 '19 at 14:25

First solve the homogeneous part $$y''-2y'+5y=0$$ by taking $$y-e^{mx}$$. Then $$m^2-2m+5=0$$ gives $$m=1\pm 2i$$. Hence we get two linearly independent solutions: $$y_1(x)=e^x \sin 2x,~~ y_2(x)=e^x \cos 2x ~~~(1)$$ Next the total solution of the required in-homogeneous ODE: $$Y''-2Y'+5Y=e^x \cos 2x=f(x) ~~~(2)$$ is given by $$Y(x)=C_1(x) y_1(x)+ C_2 y_2(x)~~~(3)$$, where by the method of variation of parameters

$$C_1(x)=-\int \frac{y_2(x) f(x)}{W(x)} dx+~~D_1, ~~~C_2(x)=\int \frac{y_1(x) f(x)}{W(x)} dx ~~+D_2~~~(4)$$

Here $$W(x)=y_1(x)y'_2(x)-y'_1 y_2(x).~~~(5)$$

Finally use (1) in (3), and (5) to get the total solution of (2), with two constants $$D_1, D_2.$$

• why do you add D1 and D2? Aug 16 '19 at 4:02
• @Meraj Haq Because ultimately $D_1 e^{x}\sin 2x +D_2 e^{x} \cos 2x$ will automatically be the general part of the solution where as rest in (3) will be without a constant and this part is called particilar integral. Aug 16 '19 at 4:18
• The particular solution is defined only up to a constant. However, notice that when you carry the constants through, you merely wind up with a restatement of the homogeneous solution. This is why they are often left off. Aug 16 '19 at 4:19
• RightHindSd Other option is not to have$D_1$ and $D_2$ in (4) and write $Y=y_g+y_p=D_1 e^x \sin 2x +D_2 e^x \cos 2x+y_p,$ Then $y_p$ will be fiven by (3) without $D_1, D_2$. Let me re-emphasize that the particular integral $y_p$ cannot have any constant. Aug 16 '19 at 4:42