Question

Solve (D2 - 4D + 4) y = x2 sin x + e2x + 3 ​

Answer 1

Answer:

5 is correct answer

Step-by-step explanation:

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Answer 2

The solution of the given differential equation is $y = (C_1 + C_2 x)e^{2x}+ \frac{1}{625} [(220x+224)cosx + (40x + 33)sinx] + \frac{x^{2} }{2} e^{2x}+ \frac{3}{4}$

Step-by-step Explanation:

Given: Differential Equation $(D^2 - 4D + 4) y = x^2 sin x + e^{2x} + 3$

To Find: Solution of the given differential equation

Solution:

• Finding Complementary Factor

Given the differential equation $(D^2 - 4D + 4) y = x^2 sin x + e^{2x} + 3$ such that consider $m^2 - 4m + 4 = 0$. Therefore, $\Rightarrow m = 2, 2$

For real and equal roots, C.F. is $(C_1 + C_2 x)e^{mx}$

Therefore, the Complementary Factor of the differential equation is

$y_{_{C.F.}} = (C_1 + C_2 x)e^{2x}$

• Finding Particular Integral

For the given differential equation $(D^2 - 4D + 4) y = x^2 sin x + e^{2x} + 3$, consider the following,

$y = \frac{1}{(D^2 - 4D + 4)} (x^2 sin x + e^{2x} + 3)$

$\Rightarrow y = \frac{1}{(D^2 - 4D + 4)} x^2 sin x + \frac{1}{(D^2 - 4D + 4)} e^{2x} + \frac{1}{(D^2 - 4D + 4)} 3$

Now, let $P.I. (1) = \frac{1}{(D^2 - 4D + 4)} x^2 sin x$

$\Rightarrow P.I. (1) = \text{imaginary part of} \frac{1}{(D^2 - 4D + 4)} x^2 e^{ix}$

$\Rightarrow P.I. (1) = \text{I.P.} e^{ix} \frac{1}{(D+i)^2 - 4(D+i) + 4} x^2 \ \ \Big[\because \frac{1}{f(D)} e^{ax} \phi(x) = e^{ax} \frac{1}{f(D+a)} \phi(x) \Big]$

$\Rightarrow P.I. (1) = \text{I.P.} e^{ix} \frac{1}{D^2 - 2D(i-2) + 4i +3} x^2$

$\Rightarrow P.I. (1) = \text{I.P.} \frac{e^{ix}}{D^{2}} \Big[{1 + (\frac{2(i-2)}{D} + \frac{4i}{D^2} + \frac{3}{D^2}) \Big]^{-1} x^2 \ \ \Big[ \because \frac{1}{f(D)} x^n = [f(D)]^{-1} x^n \Big]$

Solving the above expression by binomial expansion and extracting the imaginary part, we get;

$\Rightarrow P.I. (1) = \frac{1}{625} [(220x+224)cosx + (40x + 33)sinx]$

Now, let $P.I. (2) = \frac{1}{(D^2 - 4D + 4)} e^{2x}$

$\Rightarrow P.I. (2) = x^{2} \frac{1}{2} e^{2x} \Big[ \because \frac{1}{f(D) e^{ax}} = x^{2} \frac{1}{f''(a) } e^{ax} \ \text{if} \ f(a) = 0 \ \& \ f'(a) = 0 \Big]$

$\Rightarrow P.I. (2) = \frac{x^{2} }{2} e^{2x}$

And, let $P.I. (3) = \frac{1}{(D^2 - 4D + 4)} 3$ such that,

$\Rightarrow P.I. (3) = \frac{1}{(D^2 - 4D + 4)} 3e^{0x}$

$\Rightarrow P.I. (3) = \frac{1}{((0)^2 - 4(0) + 4)} 3e^{0x} = \frac{3}{4}$

$\Rightarrow P.I. (3) = \frac{3}{4}$

Therefore, the Particular Integral is

$y_{_{P.I.}} = P.I. (1) + P.I. (2) + P.I. (3)$

$\Rightarrow y_{_{P.I.}} = \frac{1}{625} [(220x+224)cosx + (40x + 33)sinx] + \frac{x^{2} }{2} e^{2x}+ \frac{3}{4}$

• The complete solution

The complete solution is $y = y_{_{C.F.}} + y_{_{P.I.}}$

$\Rightarrow y = (C_1 + C_2 x)e^{2x}+ \frac{1}{625} [(220x+224)cosx + (40x + 33)sinx] + \frac{x^{2} }{2} e^{2x}+ \frac{3}{4}$

Hence, the solution of the given differential equation is $y = (C_1 + C_2 x)e^{2x}+ \frac{1}{625} [(220x+224)cosx + (40x + 33)sinx] + \frac{x^{2} }{2} e^{2x}+ \frac{3}{4}$