Consider the differential equation \(\frac{{dy}}{{dx}} + \frac{y}
A. Integration factor is x<sup style="">2</sup>
B. Integration factor is x
C. The solution of the differential equation is 4xy = x<sup style="">4</sup> + 3
D. The solution of the differential equation is 4xy = y<sup style="">4</sup> – 3
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Right Answer is:
SOLUTION
Explanation:
Given:
\(\frac{{dy}}{{dx}} + \frac{y}{x} = {x^2}x = 1,\;y = 1\)
It is linear differential equation in ‘y’
\(P = \frac{1}{x}\;;Q = {x^2}\)
\(IF = {e^{\smallint \frac{1}{x}dx\;}} = {e^{\log x}} = x\)
Solution is y(IF) = ∫ Q(IF) dx
\(\Rightarrow xy = \smallint {x^3}dx = \frac{{{x^4}}}{4} + c\)
\( \Rightarrow y = \frac{{{x^3}}}{4} + \frac{c}{x}\;\;\; - - - \left( 1 \right)\)
\(x = 1,\;y = 1,\;\left( 1 \right) \Rightarrow 1 = \frac{1}{4} + c \Rightarrow c = \frac{3}{4}\)
∴ Solution is \(y = \frac{{{x^3}}}{4} + \frac{3}{{4x}}\) (or) 4xy = x4 + 3