Question

solve the differential equation(d⁴-1) y=xsinx​

Answer 1

Step-by-step explanation:

(D2+1)y=xsinx

Particular solution

y=1D2+1xsinx

=Imaginary part of1D2+1xeix

=I.P. of eix1(D+i)2+1x becausef(D)eaxg(x)=eaxf(D+a)g(x)

=I.P. of eix1(D2+2Di)x

=I.P. of eix12Di(1−Di2)x

=I.P. of eix12Di(1+Di2+D2i24+...)x

=I.P. of 12eix(1Di+12+Di4+...)x

=I.P. of 12(cosx+isinx)(−ix22+x2+i4+...)

=12{xsinx2+1−2x24cosx}

=xsinx4+1−2x28cosx

Here 18cosxis redundant since it is already covered in complimentary function.

Answer 2

Step-by-step explanation:

Given: Differential Equation $(D^{4}-1)y = x \ sin(x)$

To Find: Solution of the given Differential Equation

Solution:

• Finding complementary function C.F.

To find the complementary function (C.F.), consider differential equation $(D^{4}-1)y = x \ sin(x)$​ such that,

$\Rightarrow (m^{4}-1) = 0$

$\Rightarrow (m^{2}-1) (m^{2}+1) = 0$

$\Rightarrow (m-1) (m+1) (m^{2}+1) = 0$

$\Rightarrow m = 1, \ -1, \ \pm \ i$

Therefore, for $m = 1, \ -1, \ \pm \ i$,

$y_{_{C.F.}} = C_1 e^{x} + C_2 e^{-x} + C_3 \ cos(x) + C_4 \ sin(x) \cdots \cdots (1)$

• Finding particular integral P.I.

For particular integral (P.I.), consider $(D^{4}-1)y = x \ sin(x)$ such that,

$P.I. = \frac{1}{(D^{4}-1)} x \ sin(x) = \text{Imaginary Part (I.P.) of } e^{ix} \frac{1}{(D^{4}-1)} x$

We know that, $e^{ax} \frac{1}{f(D)} x = e^{ax} \frac{1}{f(D+a)} x$. Therefore,

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

$\Rightarrow \text{(I.P.) } e^{ix} \frac{1}{(D^{4} + i^{4} + 4D^{3}i + 6D^{2}i^{2} +4Di^{3} -1)} x$

$\Rightarrow \text{(I.P.) } e^{ix} \frac{1}{(D^{4} + 4D^{3}i - 6D^{2} -4Di)} x$

$\Rightarrow \text{(I.P.) } \frac{e^{ix}}{-4iD} [\frac{1}{1 + \frac{(D^{4} + 4D^{3}i - 6D^{2})}{4iD}}] x$

$\Rightarrow \text{(I.P.) } \frac{e^{ix}}{-4iD} [{1 - \frac{(D^{4} + 4D^{3}i - 6D^{2})}{4iD}}]^{-1} x$

Now, expand the term $[{1 - \frac{(D^{4} + 4D^{3}i - 6D^{2})}{4iD}}]^{-1}$ further by binomial expansion to get,

$\Rightarrow \text{(I.P.) } \frac{ie^{ix}}{4D} [{1 - \frac{(D^{3} + 4D^{2}i - 6D)}{4i}} + \frac{(D^{3} + 4D^{2}i - 6D)^{2} }{16i^{2}}}] x$

$\Rightarrow \text{(I.P.) } \frac{ie^{ix}}{4} [ \frac{1}{D} + \frac{13}{4}D - \frac{3}{2i}] x$

$\Rightarrow \text{(I.P.) } \frac{e^{ix}}{4} [ \frac{x^{2}i}{2} + \frac{13}{4}i - \frac{3x}{2}]$

Since $e^{ix} = cos(x) + i sin(x)$, therefore, we can write;

$\Rightarrow \text{(I.P.) } \frac{cos(x) + i sin(x)}{4} [ \frac{x^{2}i}{2} + \frac{13}{4}i - \frac{3x}{2}]$

$\Rightarrow \text{(I.P.) } \frac{1}{4} [ \frac{(cos(x))x^{2}i}{2} + \frac{13}{4} (cos(x))i - \frac{3x}{2} (cos(x)) + \frac{(sin(x))x^{2}(-1)}{2} + \frac{13}{4} (sin(x))(-1) - \frac{3x}{2} (sin(x))i]$

The P.I. of the differential equation is the imaginary part of the above expression, thus we can write,

$y_{_{P.I.}} = \frac{13}{16} cos(x)+ \frac{1}{8}[x^{2} cos(x) - 3xsin(x)] \cdots \cdots (2)$

• Complete solution of the differential equation

The complete solution of the given differential equation is,

$y = y_{_{C.F.}} + y_{_{P.I.}}$

$\Rightarrow y = C_1 e^{x} + C_2 e^{-x} + (C_3 + \frac{13}{16}) \ cos(x) + C_4 \ sin(x) + \frac{1}{8}[x^{2} cos(x) - 3xsin(x)]$

Hence, the complete solution of the differential equation is

$y = C_1 e^{x} + C_2 e^{-x} + (C_3 + \frac{13}{16}) \ cos(x) + C_4 \ sin(x) + \frac{1}{8}[x^{2} cos(x) - 3xsin(x)]$