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Question: How to solve \(\dfrac{{{d^2}y}}{{d{x^2}}} + {a^2}y = 0\)....

How to solve d2ydx2+a2y=0\dfrac{{{d^2}y}}{{d{x^2}}} + {a^2}y = 0.

Explanation

Solution

To solve this problem we will use the method of undetermined coefficients. In this method, we take a second order differential equation and in this case it is a homogeneous equation. Then, we will convert the equation into its characteristic equation and then accordingly solve the equation. Then we will write the solution of the differential equation based on the solution of the characteristic equation. So, let us see how to solve the problem.

Complete solution:-
The given differential equation is,
d2ydx2+a2y=0\dfrac{{{d^2}y}}{{d{x^2}}} + {a^2}y = 0
Now, the characteristic equation of the differential equation is,
r2+a2=0{r^2} + {a^2} = 0
Now, subtracting both sides by a2{a^2}, we get,
⇒r2=−a2\Rightarrow {r^2} = - {a^2}
⇒r2=−1.a2\Rightarrow {r^2} = - 1.{a^2}
Taking square root on both the sides, we get,
⇒r=±−1.a\Rightarrow r = \pm \sqrt { - 1} .a
We know, −1=i\sqrt { - 1} = i.
So, using this property, we get,
⇒r=±ai\Rightarrow r = \pm ai
⇒r=ai or r=−ai\Rightarrow r = ai{\text{ or }}r = - ai
Therefore, by the method of undetermined coefficients, the solution of the differential equation, will be,
y=c1eaix+c2e−aixy = {c_1}{e^{aix}} + {c_2}{e^{ - aix}}
Now, we know, eiθ=cos⁡θ+isin⁡θ{e^{i\theta }} = \cos \theta + i\sin \theta and e−iθ=cos⁡θ−isin⁡θ{e^{ - i\theta }} = \cos \theta - i\sin \theta .
Thus, by using this property, we get,
⇒y=c1(cos⁡ax+isin⁡ax)+c2(cos⁡ax−isin⁡ax)\Rightarrow y = {c_1}\left( {\cos ax + i\sin ax} \right) + {c_2}\left( {\cos ax - i\sin ax} \right)
⇒y=c1cos⁡ax+ic1sin⁡ax+c2cos⁡ax−ic2sin⁡ax\Rightarrow y = {c_1}\cos ax + i{c_1}\sin ax + {c_2}\cos ax - i{c_2}\sin ax
Taking the cosine and sine terms common, we get,
⇒y=cos⁡ax(c1+c2)+sin⁡ax(ic1−ic2)\Rightarrow y = \cos ax\left( {{c_1} + {c_2}} \right) + \sin ax\left( {i{c_1} - i{c_2}} \right)
Now, here, c1+c2=A{c_1} + {c_2} = A, since, sum of constants is also a constant.
And, ic1−ic2=Bi{c_1} - i{c_2} = B, since, subtracting a constant from another constant is also a constant.
⇒y=Acos⁡ax+Bsin⁡ax\Rightarrow y = A\cos ax + B\sin ax

Therefore, the solution of the differential equation d2ydx2+a2y=0\dfrac{{{d^2}y}}{{d{x^2}}} + {a^2}y = 0 is y=Acos⁡ax+Bsin⁡axy = A\cos ax + B\sin ax.

Notes:-
There is also an alternative method to this problem. We can also reduce this second order differential equation into a first order differential equation, by assuming
dydx=p\dfrac{{dy}}{{dx}} = p
⇒d2ydx2=dpdx\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{dp}}{{dx}}
⇒d2ydx2=dpdy.dydx\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = \dfrac{{dp}}{{dy}}.\dfrac{{dy}}{{dx}}
⇒d2ydx2=pdpdy\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} = p\dfrac{{dp}}{{dy}}
Thus, by substituting these values, we can solve the equation as a first order differential equation and find the required answer.