How to Get the Solution of the Differential Equation

1

0=+ xkxm tBtAx nn ωω cossin +=in the Form:

or ?

FAQ 1

( )φω += tXx nsin

2

• The differential equation

is a second order homogeneous differential equation with constant coefficients. (k and m are constants.)

• Since the coefficients are constants, we can assume solution to Eq. (1) in the exponential form

where s is a constant yet to be determined.

0=+ xkxm )1(

)2(tsex =)3(tsesx =∴

)4(2 tsesx =∴

3

• Substituting for in Eq. (1) , we get the characteristic equation as

• The two roots of the characteristic equations are:

• Letting , the two roots are:

02 =+ tsts ekesm

xx and

)5(02 =+ ksm

)6(12,1 mki

mk

mks ±=−±=−±=

nn isis ωω −=+= 21 and

mkn /=ω

• Substituting these roots in Eq. (2), we get two possible solutions:

• These are two linearly independent solutions to Eq. (1). So, any linear combination of them will also be a solution. The general solution can therefore be written as

• The constants C1 and C2 can be determined in terms of initial conditions., viz.,

• The solution given by Eq. (8) is in exponential form. Of course, we can stop here as we have already got the general solution to the differential equation. However, next slide shows a more convenient form of solution. 4

)7(2

1

==

−

+

ti

ti

n

n

eCxeCx

ω

ω

)8(21titi nn eCeCx ωω −+ +=

00 )0( and )0( xtxxtx ====

5

• Using Euler’s formula , we can write

• Substituting Eq. (9) and (10) in Eq. (8) and rearraning,

• Since C1 and C2 are arbitrary constants, we can rewrite Eq. (11) using another set of constants A and B as†

• The constants A and B can be determined based on initial conditions. Eq. (12) is in trigonometric form.

• Next slide shows another convenient form of solution.

( ) ( ) )11(sincos 2121 tCCitCCx nn ωω −++=

)12(cossin tBtAx nn ωω +=

xixeix sincos +=

)9(sincos tite nnti n ωωω +=

)10(sincos tite nnti n ωωω −=−

† For real values of initial displacement and initial velocity, i(C1-C2) and (C1+C2) will both turn out to be real. So, we can represent them by real constants A and B .

6

• We can also rewrite Eq. (12) in an equivalent form as

• The constants X and φ can be determined based on initial conditions.

• Sine Eq. (12) and (13) are equivalent, we can now find how the constants X and φ are related to the constants A and B. In order to do that:

• Expand Eq. (13) as

( ) )13(sin φω += tXx n

( ) )14(sincoscossin φωφω ttXx nn +=

7

• Comparing the coefficients of and on the right sides of Eqs. (12) and (14):

• Squaring both sides of Eq. (15) & Eq. (16) and adding, we get

• Dividing Eq. (16) by Eq(15), we get

• So Eq. (13) can be written as

)16(sin BX =φ)15(cos AX =φ

)17(22 BAX +=

)19(tan 1

= −

ABφ

tnωsin tnωcos

( )( ) )20(tansin 122AB

n tBAx −++= ω