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Question

The solution of the differential equation $fraction numerator d y over denominator d x end fraction plus fraction numerator 3 x to the power of 2 end exponent over denominator 1 plus x to the power of 3 end exponent end fraction y equals fraction numerator s i n to the power of 2 end exponent x over denominator 1 plus x to the power of 3 end exponent end fraction$ is

$y (1 + x^{3}) = x + \frac{1}{2} s i n 2 x + c$

$y (1 + x^{3}) = c x + \frac{1}{2} s i n 2 x$

$y (1 + x^{3}) = c x - \frac{1}{2} s i n 2 x$
$y (1 + x^{3}) = \frac{x}{2} - \frac{1}{4} s i n 2 x + c$

hint Hint:

We are given a differential equation. We have to find the solution of the differential equation. For this question, we will use the method of finding integrating factor. Integrating factor makes the differential equation simple for integration.

The correct answer is: $y left parenthesis 1 plus x to the power of 3 end exponent right parenthesis equals fraction numerator x over denominator 2 end fraction minus fraction numerator 1 over denominator 4 end fraction blank s i n blank 2 x plus c$

The given equation is of type $fraction numerator d y over denominator d x end fraction space plus P y space equals space Q space space$ .
Here P and Q can be function of x or constants.
So, we will use the method of finding integrating factor.
Integrating factor is given by.
$I. F space equals space e to the power of integral P d x end exponent$
The given equation is $fraction numerator d y over denominator d x end fraction plus fraction numerator 3 x squared over denominator 1 plus x cubed end fraction y space equals space fraction numerator sin squared x over denominator 1 plus x cubed end fraction$
Comparing with the above equation $P space equals space fraction numerator 3 x squared over denominator 1 space plus x cubed end fraction$
Substituting the values of P, in the integrating factor.
$I. F space equals e to the power of integral fraction numerator 3 x squared over denominator 1 space plus space x cubed end fraction d x end exponent I f space w e space s e e space t h e space n u m e r a t o r space o f space i n t e g r a l space o r space P space i s space d e r i v a t i v e space o f space t h e space d e n o m i n a t o r. I space equals integral fraction numerator f apostrophe left parenthesis x right parenthesis over denominator f left parenthesis x right parenthesis end fraction d x space equals log f left parenthesis x right parenthesis I. F space equals e to the power of log left parenthesis 1 space plus space x cubed right parenthesis end exponent space space space space space space space space space space space space space equals 1 space plus space x cubed space space space space space space space space space space space space space space space space space... left parenthesis e to the power of log subscript e a space end exponent equals space a right parenthesis$
So, I.F = 1 + x³
Now, the solution of such equation is given by
$y left parenthesis I. F right parenthesis space equals space integral Q left parenthesis I. F right parenthesis d x space plus space C$
$y left parenthesis 1 space plus x cubed right parenthesis space equals integral fraction numerator s i n squared x over denominator 1 space plus space x cubed end fraction left parenthesis 1 space plus x cubed right parenthesis space d x space plus space C space space space space space space space space space space space space space space space space equals integral s i n squared x space d x space plus space C space space space space space space space space space space space space space space space space equals integral fraction numerator 1 space minus space cos 2 x over denominator 2 end fraction d x space plus space C space space space... left parenthesis f o r m u l a space o f space C o s 2 x right parenthesis space space space space space space space space space space space space space space space space equals integral left parenthesis 1 half space minus fraction numerator cos 2 x over denominator 2 end fraction right parenthesis d x space plus space C space y left parenthesis 1 space plus x cubed right parenthesis space equals space x over 2 space minus space fraction numerator sin 2 x over denominator 4 end fraction plus space C$
This is the required solution of the differential equation.

For such questions, we should know different methods to solve differential equation.

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