Solve the Equation d^2y/dx^2 + 4y = cos2x | (D^2+4)y=cos2x

Here we solve the equation d^2y/dx^2+4y=cos2x, that is, Solve (D^2+4)y=cos2x. The general solution of d²y/dx² + 4y = cos2x is equal to y=c₁cos2x + c₂sin2x + (xsin2x)/4.

Solve (D²+4)y=cos2x

Question: Solve the equation $\dfrac{d^2y}{dx^2}$ + 4y = cos2x.

Solution:

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

(D²+4)y = cos2x

To solve the equation d²y/dx² + 4y = cos2x, we need to find the complementary function (CF) and particular integral (PI).

[Complementary Function (CF)]:

The corresponding auxiliary equation is given by

m²+4 = 0

This implies, 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)]:

The particular integral is given by

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

= $\dfrac{1}{-2^2+4}\cos 2x$, replacing D² = -a² = -2². So this formula fails because -2²+4 = 0.

Therefore,

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

= $x \dfrac{1}{\frac{d}{dD}(D^2+4)}\cos 2x$

= $x \dfrac{1}{2D}\cos 2x$

= $\dfrac{x}{2} \dfrac{1}{D}\cos 2x$

= $\dfrac{x}{2} \displaystyle \int \cos 2x ~dx$

= $\dfrac{x}{2} \dfrac{\sin 2x}{2}$

= $\dfrac{x\sin 2x}{4}$

∴ PI = $\dfrac{x\sin 2x}{4}$.

General Solution: Hence, the general solution of $\dfrac{d^2y}{dx^2} + 4y = \cos 2x$ is equal to

y = CF + PI

⇒ y = c₁cos2x + c₂sin2x $+ \dfrac{x \sin 2x}{4}$ where c₁ and c₂ are arbitrary constants.

Also Study:

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