How to Solve (D^2+4)y=cos^2x

The general solution of (D²+4)y=cos²x is given by y=c₁cos2x + c₂sin2x + (1+xsin2x)/8. Lets learn how to solve (D^2+4)y=cos^2x. As D² = d²/dx², the differential equation (D²+4)y=cos²x can also be written as follows: d²y/dx² + 4y = cos²x.

Solve (D²+4)y=cos^2x

Question: Solve the equation $\dfrac{d^2y}{dx^2} + 4y = \cos^2x$.

Solution:

Letting D² = d²/dx², we can rewrite the given differential equation as follows:

(D²+4)y = cos²x

[Complementary Function (CF)]:

The auxiliary equation is given by the solution of (D²+4)y = 0.

Let y=e^mx be a solution.

Then Dy = dy/dx = me^mx and D²y = d²y/dx² = m²e^mx

So from (D²+4)y = 0, we obtain that

(m²+4)e^mx = 0

As e^mx is never 0, we must have that

m²+4 = 0

⇒ m² = -4

Taking square root on both sides,

m = $\pm \sqrt{-4}$

⇒ m = $\pm 2i$

Therefore, CF = c₁cos2x + c₂sin2x where c₁ and c₂ are arbitrary constants.

[Particular Integral (PI)]:

Using the following trigonometric identity 2cos²x = 1+cos2x ⇒ cos²x = $\dfrac{1+\cos 2x}{2}$, the particular integral is given by

$\dfrac{1}{D^2+4}\cos^2x$

= $\dfrac{1}{2} \dfrac{1}{D^2+4}(1+\cos2x)$

= $\dfrac{1}{2} \dfrac{1}{D^2+4} 1+ \dfrac{1}{2} \dfrac{1}{D^2+4} \cos2x$

Note: Replacing D2 = -a2 = -22, we have D2+4 = -22+4 = 0. Thus we cannot compute 1/(D2+4) cos2x by usual method.

= $\dfrac{1}{2\cdot 4} \left( 1+\dfrac{D^2}{4} \right)^{-1} 1$ $+ \dfrac{1}{2} \cdot x \dfrac{1}{\frac{d}{dD}(D^2+4)} \cos2x$

= $\dfrac{1}{8} \left( 1-\dfrac{D^2}{4}+(\dfrac{D^2}{4})^2-\cdots \right) 1$ $+ \dfrac{1}{2} \cdot x\dfrac{1}{2D} \cos2x$

= $\dfrac{1}{8}+ \dfrac{x}{4}\dfrac{1}{D} \cos2x$

= $\dfrac{1}{8} + \dfrac{x}{4}\displaystyle \int \cos2x~dx$

= $\dfrac{1}{8} + \dfrac{x \sin 2x}{8}$

= $\dfrac{1}{8} + \dfrac{x \sin 2x}{8}$

∴ Particular Integral of (D²+4)y=cos²x is equal to PI = $\dfrac{1}{8} + \dfrac{x \sin 2x}{8}$.

General Solution: So the general solution of (D²+4)y=cos²x is given by

y = CF + PI

⇒ y = c₁cos2x + c₂sin2x $+ \dfrac{1+x \sin 2x}{8}$.

Hence, y=c₁cos2x + c₂sin2x + (1+xsin2x)/8 is the solution of (D²+4)y=cos²x, where c₁ and c₂ are arbitrary constants of integrals.

Also Study:

Solve (D2+4)y=sin2x	Solve (D2+4)y=sin2x
Solve (D2+4)y=cos2x	Solve dy/dx = ex+y
Solve dy/dx = sin(x+y)	Solve dy/dx = cos(x+y)

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