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

Here we learn how to solve (D^2+4)y=sin^2x. The general solution of (D²+4)y=sin²x is given by y=c₁cos2x + c₂sin2x + (1-xsin2x)/8.

The differential equation (D²+4)y=sin²x can also be written as follows: d²y/dx² + 4y = sin²x. Now let us solve this equation.

Solve (D²+4)y=sin^2x

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

Solution:

One can rewrite the given differential equation as follows:

(D²+4)y = sin²x

where D² = d²/dx².

[Complementary Function (CF)]:

For the complementary function, the auxiliary equation is given by

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)]:

Note that 2sin²x = 1-cos2x.

So, sin²x = $\dfrac{1-\cos 2x}{2}$.

The particular integral is given by

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

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

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

Note: As D² = -a² = -2², we have D²+4 = -2²+4 = 0, and so 1/(D²+4) cos2x cannot be computed by usual method. The above will be equal to

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

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

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

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

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

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

∴ Particular Integral of (D²+4)y=sin²x is equal to PI = $\dfrac{1}{8} – \dfrac{x \sin 2x}{8}$.

General Solution: So the general solution of (D²+4)y=sin²x 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=sin²x, where c₁ and c₂ are arbitrary constants of integrals.

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