The solution of (D$^2$ + 16) y = cos4x is
- Acos4x + Bsin 4x + $\frac{x}{8}sin 4x$
- Acos4x + Bsin 4x - $\frac{x}{8}sin 4x$
- Acos4x + Bsin 4x + $\frac{x}{4}sin 4x$
- Acos4x + Bsin 4x - $\frac{x}{4}sin 4x$
The Correct Option is A
Approach Solution - 1
If (D$^2$ + 16)y = cos 4x
Here the auxiliary equation is m$^2$ + 16 = 0
$\Rightarrow$ m = 4
$\therefore$ Complementary function = (A cos 4x + B sin 4x) &
Particular Integral (P.I.) $= \frac{1}{D^2 + 16}. \, cos4x$
But $\frac{1}{D^2 \, + \, a^2} cos ax = \frac{x}{2a}sin ax$
$\therefore$ P.I.=$\frac{x}{2 \times 4}$. sin 4x
= $\frac{x}{8}sin4x$
$\therefore $ Solution y = Complementary function $+$ Particular Integral
$\Rightarrow $ y = A cos 4x + B sin 4x + $\frac{x}{8}$sin 4 x
Approach Solution -2
P.I. = \(\frac{cos\,4x}{D^2+16}\)
\(= \frac{cos\,4x}{D^2+4^2}\)\(=\frac{x\,sin\,4x}{8}\)
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Concepts Used:
Integral
The representation of the area of a region under a curve is called to be as integral. The actual value of an integral can be acquired (approximately) by drawing rectangles.
- The definite integral of a function can be shown as the area of the region bounded by its graph of the given function between two points in the line.
- The area of a region is found by splitting it into thin vertical rectangles and applying the lower and the upper limits, the area of the region is summarized.
- An integral of a function over an interval on which the integral is described.
Also, F(x) is known to be a Newton-Leibnitz integral or antiderivative or primitive of a function f(x) on an interval I.
F'(x) = f(x)
For every value of x = I.
Types of Integrals:
Integral calculus helps to resolve two major types of problems:
- The problem of getting a function if its derivative is given.
- The problem of getting the area bounded by the graph of a function under given situations.