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Question Number 105877 by john santu last updated on 01/Aug/20

	Answered by bemath last updated on 01/Aug/20
	$(dy/dx).(1/x^3 ) + ((1+y^2 )/(y^4 (1+x^2 ))) = 0 ((y^4 dy)/(1+y^2 )) = −((x^3 dx)/(1+x^2 )) ∫ ((y^4 dy)/(1+y^2 )) = −∫ ((x^3 dx)/(1+x^2 )) let y = tan p and x = tan q ∫ ((tan^4 p sec^2 p dp)/(sec^2 p)) = −∫((tan^3 q sec^2 q dq)/(sec^2 q)) ∫ tan^4 p dp = −∫ tan^3 q dq ∫ tan^2 p (sec^2 p−1)dp = −∫tan q(sec^2 q−1)dq I_1 = ∫ tan^2 p d(tan p)−∫d(tan p)+p = (1/3)tan^3 p−tan p+p = (1/3)y^3 −y+ tan^(−1) (y) I_2 = −{∫tan q d(tan q)−∫ tan q dq} I_2 =−(1/2)tan^2 q + ln ∣cos q∣ I_2 =−(1/2)x^2 + ln ∣(1/(√(1+x^2 )))∣ ∴ General solution (1/3)y^3 −y+ tan^(−1) (y)=−(1/2)x^2 −ln(√(1+x^2 )) +C$
	$\frac{d y}{d x} . \frac{1}{x^{3}} + \frac{1 + y^{2}}{y^{4} (1 + x^{2})} = 0$ $\frac{y^{4} d y}{1 + y^{2}} = - \frac{x^{3} d x}{1 + x^{2}}$ $\int \frac{y^{4} d y}{1 + y^{2}} = - \int \frac{x^{3} d x}{1 + x^{2}}$ $l e t y = \tan p a n d x = \tan q$ $\int \frac{\tan^{4} p \sec^{2} p d p}{\sec^{2} p} = - \int \frac{\tan^{3} q \sec^{2} q d q}{\sec^{2} q}$ $\int \tan^{4} p d p = - \int \tan^{3} q d q$ $\int \tan^{2} p (\sec^{2} p - 1) d p = - \int \tan q (\sec^{2} q - 1) d q$ $I_{1} = \int \tan^{2} p d (\tan p) - \int d (\tan p) + p$ $= \frac{1}{3} \tan^{3} p - \tan p + p$ $= \frac{1}{3} y^{3} - y + \tan^{- 1} (y)$ $I_{2} = - {\int \tan q d (\tan q) - \int \tan q d q}$ $I_{2} = - \frac{1}{2} \tan^{2} q + \ln ∣ \cos q ∣$ $I_{2} = - \frac{1}{2} x^{2} + \ln ∣ \frac{1}{\sqrt{1 + x^{2}}} ∣$ $∴ G e n e r a l s o l u t i o n$ $\frac{1}{3} y^{3} - y + \tan^{- 1} (y) = - \frac{1}{2} x^{2} - \ln \sqrt{1 + x^{2}} + C$
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