Practicing Success
Practicing Success
CUET Preparation Today
Find the general solution of the DE : \(\frac{dy}{dx}\) = \(\frac{1+y^2}{1+x^2}\) |
\(\tan^{-1} {x}\) = \(\tan^{-1} {y}\) + C \(\tan^{-1} {x}\) - \(\tan^{-1} {y}\) = C \(\tan^{-1} {y}\) = \(\tan^{-1} {x}\) + C All the above |
All the above |
\(\frac{dy}{dx}\) = \(\frac{1+y^2}{1+x^2}\) dx . 1+x2 = dy . 1+y2 and then integrate to get : \(\tan^{-1} {x}\) = \(\tan^{-1} {y}\) + C , or \(\tan^{-1} {y}\) = \(\tan^{-1} {x}\) + C, or \(\tan^{-1} {x}\) - \(\tan^{-1} {y}\) = C Note : -ve sign can be incorporated into the constant C
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