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If dy/dx+ (1-y^2/1-x^2), what is the equation corresponding to the differential equation? Answer Follow · 1 Request


one month ago

							or, \frac{dy}{dx\ }\ +\ \frac{1-y^{2}}{1-x^{2}}=0or,\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

one month ago
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