Question

Solve x2D
2
-3xD+4y = x2
cos(logx)

Answer 1

Answer:

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Answer 2

The solution of the differential equation is $y = x^{2} (C_1 + C_2 (logx) - cos(logx) )$

Step-by-step explanation:

Given: $(x^{2} D^{2} - 3xD + 4)y = x^{2} cos(logx)$

To find: Solution of the differential equation

Solution:

• Finding Complementary Function (C.F.)

For the given differential equation, let $x=e^z$ such that $z = logx$ and $D \equiv \frac{d}{dz}$  such that $x \frac{dy}{dx} = Dy$ and $x^{2} \frac{d^{2}y}{dx^{2} } = D(D-1)y$

Thus; the equation becomes,

$\Rightarrow D(D-1)y - 3Dy + 4y = e^{2z} cos(z)$

$\Rightarrow (D-2)^{2}y = e^{2z} cos(z) \ \cdots \cdots (1)$

Now, considering (1) such that, $\Rightarrow (m-2)^{2} = 0$ $\Rightarrow m = 2, 2$

Since, the roots are real and equal, then,

$\Rightarrow y_{C.F.} = (C_1 + C_2 z)e^{2z}$

$\Rightarrow y_{C.F.} = (C_1 + C_2 (logx))x^{2} \ \cdots \cdots (2)$

• Finding Particular Integral (P.I.)

To find the Particular Integral, consider the following:-

$y_{P.I.} = \frac{1}{D^{2} -4D + 4}e^{2z}cosz$

$\Rightarrow y_{P.I.} = e^{2z} \frac{1}{(D+2)^{2} -4(D+2) + 4}cosz \ \ \ \ \because \frac{1}{f(D)} e^{az} cosz = e^{az} \frac{1}{f(D+a)} cosz$

$\Rightarrow y_{P.I.} = e^{2z} \frac{1}{D^{2}}cosz$

$\Rightarrow y_{P.I.} = - e^{2z} cosz \ \ \ \ \because \frac{1}{D} = \int$

$\Rightarrow y_{P.I.} = - x^{2} cos(logx) \ \cdots \cdots (3)$

• Solution of the differential equation

Complete solution of the given differential equation is $y = y_{C.F.} + y_{P.I.}$

From (3) and (4),

$y = (C_1 + C_2 (logx))x^{2} - cos(logx) x^{2} = x^{2} (C_1 + C_2 (logx) - cos(logx) )$

Hence, the solution of the differential equation is $y = x^{2} (C_1 + C_2 (logx) - cos(logx) )$