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If dy/dx+ (1-y^2/1-x^2), what is the equation corresponding to the differential equation?
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rekha , 4 Years ago
Grade 12
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Srutarshi Tripathi

Last Activity: 4 Years ago

or, \frac{dy}{dx\ }\ +\ \frac{1-y^{2}}{1-x^{2}}=0
or,\frac{dy}{dx}=\ -\frac{\left(1-y^{2}\right)}{1-x^{2}}
or, \frac{dy}{1-y^{2}}=-\ \frac{dx}{1-x^{2}}
or, \int_{ }^{ }\frac{dy}{1-y^{2}}=-\int_{ }^{ }\frac{dx}{1-x^{2}}
or, \ln\left(\frac{\left(1+y\right)}{1-y}\right)=-\ln\left(\frac{\left(1+x\right)}{1-x}\right)-\ \ln\left(k\right)  (ln(k) is constant of integration)
or, \frac{\left(1+y\right)}{1-y}\cdot\frac{\left(1+x\right)}{1-x}\cdot k=1
 

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