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 July 14th, 2013, 06:42 AM #1 Senior Member   Joined: Sep 2012 Posts: 112 Thanks: 0 second order differential equation Could you please explain(in details) the calculation of complementary function and particular integral of the following function step by step? $y^'^' + 5y^'= x^2 + 7x + 5$
July 14th, 2013, 07:55 AM   #2
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Re: second order differential equation

Quote:
 Originally Posted by helloprajna Could you please explain(in details) the calculation of complementary function and particular integral of the following function step by step? $y^'^' + 5y^'= x^2 + 7x + 5$
so to solve this first consider the homogenous case (RHS is zero)

$y'' + 5y' = 0$

you make an educated guess that the solution will be of the form, $y=e^{\lambda x}$ for some constant $\lambda$ which we will determine.

so differentiate this:

$y'=\lambda e^{\lambda x} \\
y'' =\lambda^2 e^{\lambda x}$

and substituting in these values to the homogeneous equation

$\lambda^2 e^{\lambda x} + 5(\lambda e^{\lambda x})= 0$

factor out,
$(\lambda^2 + 5\lambda)e^{\lambda x}=0$

since $e^{\lambda x} \neq 0, \forall x$ we can cancel this factor

giving the auxillary equation:

$\lambda^2 + 5\lambda=0$

so we have, $\lambda=0$ and $\lambda=-5$

Giving the complementary function: $y=C_1+ C_2e^{-5x}$

For the particular integral, try a solution of the form:

$y=Ax^3+Bx^2+Cx$
DIfferentiating:

$y'=3Ax^2+2Bx+C$
$y''=6Ax+2B$

And subbing into original equation, we get:

$6Ax+2B +5(3Ax^2+2Bx+C)=x^2+7x+5$

compare coefficients to get the values for A, B and C

$15A=1 \Rightarrow A=\frac{1}{15}$
$6A +10B=7 \Rightarrow B=\frac{33}{50}$
$2B +5C=5 \Rightarrow C= \frac{92}{125}$

The general solution is the complementary function plus the particular integral

$y=C_1+ C_2e^{-5x}+ \frac{1}{15}x^3+ \frac{33}{50}x^2 + \frac{92}{125}x$

with $C_1$and$C_2$ determined by the initial conditions

 July 14th, 2013, 09:39 PM #3 Senior Member   Joined: Sep 2012 Posts: 112 Thanks: 0 Re: second order differential equation Excellent explanation! I've just one question. According to my book if $f(x)= ax^2$, then I should put particular integral=$Ax^2 + Bx + C$. In this problem, this isn't working. So I have to multiply x with$Ax^2 + Bx + C$, like you have shown to me. But in my answer I must explain why do I multiply x with $Ax^2 + Bx + C$. So what is the argument/logic behind taking particular integral$= x(Ax^2 + Bx + C) = Ax^3 + Bx^2 + Cx$?
 July 14th, 2013, 10:16 PM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: second order differential equation You cannot have any term in the particular solution that is a solution to the corresponding homogeneous equation. The annihilator method can be used to demonstrate that this is the case. Observe that the differential operator: $A\equiv D^3$ annihilates: $g(x)=x^2+7x+5$ Applying this operator to the given ODE, we find: $D^3$$D^2+5D$$y=0$ $D^4(D+5)y=0$ Hence, the characteristic roots are: $r=-5,0$ with $r=0$ of multiplicity 4, hence the general solution will take the form: $y(x)=c_1e^{-5x}+c_2+c_3x+c_4x^2+c_5x^3$ We know though, that for the original ODE, the corresponding homogeneous solution is: $y_h(x)=c_1e^{-5x}+c_2$ and by the principle of superposition, we are left to conclude that there must exist a particular solution of the form: $y_p(x)=c_3x+c_4x^2+c_5x^3$
 July 15th, 2013, 07:43 AM #5 Senior Member   Joined: Sep 2012 Posts: 112 Thanks: 0 Re: second order differential equation I tried my best Mark, but I couldn't understand your explanation. I am giving below an example to show what I exactly want to know. In a problem of my book, the complementary function is $Ce^2^x + C'e^3^x$ and $g(x)= e^2^x$. I write in my answer paper that as a rule I should put particular integral= $Ae^2^x$. But I won't do it because if I put C= A & C'=0 in the coplementary function, I will get $Ae^2^x$, which is the same as the particular integral. So I will put instead $Axe^2^x$ as the particular integral. I must write this type of explanation in my answer paper for the problem I've posted here. So could you please help me?
 July 15th, 2013, 08:06 AM #6 Senior Member   Joined: Aug 2011 Posts: 334 Thanks: 8 Re: second order differential equation Why not z=y' z'+5z = x²+7x+5 Solve the first order equation to obtain z(x). Then, integrate z(x) to obtain y(x).
 July 15th, 2013, 08:58 AM #7 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: second order differential equation To solve a linear equation with constant coefficients, you start by finding the "characteristic equation", the polynomial replacing the nth derivative by the nth power of r. Here that is $r^2+ 5r= r(r+ 5)= 0$ which has r= 0 and r= -5 as solutions. Then the general solution to the differential equation is a linear combination of e to those powers. Then general solution to the associated homogeneous equation here is $Ce^{0x}+ De^{-5x}= C+ De^{-5x}$ (since e to the 0 power is 1). You can also get complex number solutions. If a+ bi is a solution, then $e^a(Ccos(bx)+ Dsin(bx))$. And if there are "multiple roots" we multiply by "x" or powers of x. That is, all possible solutions of "linear differential equations with constant coefficients are (1) powers of x (including the 0 power- a constant) (2) powers of x times exponentials (3) powers of x times sine or cosine. I say that because in order to use the method of "undetermined coefficients" we have to first "guess" the "form" of the particular integral. And we can only reasonably do that when the "right side" is one of those kinds of solutions. Here, the right side is [tex]x^2+ 7x+ 5[/itex] a polynomial or "powers of x". Since powers of x are "x times a constant", that corresponds to $e^0$ which is already a solution to the homogeneous equation. Trying $Ax^2+ Bx+ C$ will not work because that "$C= C(1)= Ce^{0x}$" will give 0 when put into the equation for any value of C. So we multiply by x: try $y= Ax^3+ Bx^2+ Cx$. But it is important to note that this method only works when the "right side" function is one of those three types of functions that we can find as solutions to homogeneous equations with constant coefficients. A methods that will work for all functions on the right hand side is "variation of parameters". Here, since we already have 1 and $e^{-5x}$ as solutions to the associated homogeneous equation, we look for a solution of the form $y(x)= u(x)(1)+ v(x)e^{-5x}= u+ ve^{-5x}$. (We are replacing the constant coefficients in the general solution with variable functions- hence the name "variation of parameters".) Differentiating, $y'= u'+ v'e^{-5x}- 5ve^{-5x}$. There are, in fact, an infinite number of functions, u and v, that will work here- we can "narrow the search" and simplify by requiring that $u'+ v'e^{-5x}= 0$. That leaves only $y'= -5ve^{-5x}$. Now $y''= -5v'e^{-5x}+ 25ve^{-5x}$. Putting those into the differential equation, $y''+ 5y'= -5v'e^{-5x}+ 25ve^{-25}x- 25ve^{-25x}= -5v'e^{-5x}= x^2+ 7x+ 5$. Notice that there is no second derivative because our requirement that $u'+ v'e^{-5x}= 0$ left no first derivative to differentiate. And there is no "u" or "v" without a derivative because their coefficents, 1 and $e^{-5x}$, satisfy the homogeneous equation. The result of all that is that we have two equations, $u'+ v'e^{-5x}= 0$ and $-5v'e^{-5x}= x^2+ 7x+ 5$, to solve for u' and v'. In fact, the second equation has only v': $-5v'e^{-5x}= x^2+ 7x+ 5$ so that $v'= -(x^2+ 7x+ 5)e^{5x}/5$ which can be integrated "by parts". Also, $u'+ v'e^{-5x}= 0$ gives immediately $u'= -v'e^{-5x}$.
 July 15th, 2013, 09:26 PM #8 Senior Member   Joined: Sep 2012 Posts: 112 Thanks: 0 Re: second order differential equation Finally I got my answer. Thank you so much everyone. I am grateful to you all.

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given y=x is one solution to the corresponding homogeneous equation, find the general solution to the differential equation x^2y''-xy' y=x.

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