Question

the equation of curve satisfying differential equation
$(1 + {y}^{2} )( {e}^{x} + 1)dy = {e}^{x} + {y}^{2} dx \: and \: also \: passes \: through \: the \: point \: (0 \: 1) \: is$
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Answer 1

The equation of curve satisfying differential equation (1 + y²)(e^x + 1)dy = e^x y² dx and also passes through the point (0, 1) is ...

solution : differential equation is, (1 + y²)(e^x + 1)dy = e^x y² dx

⇒(1 + y²)/y² dy = e^x/(1 + e^x) dx

⇒∫(1/y² + 1)dy = ∫e^x/(1 + e^x) dx

⇒-1/y + y = ln(1 + e^x) + C

at (0, 1)

⇒-1/1 + 1 = ln(1 + e⁰) + C

⇒0 = ln(2) + C

⇒C = -ln(2)

now equation is , -1/y + y = ln(1 + e^x) - ln2

⇒(y² - 1)/y = ln[(1 + e^x)/2]

⇒y² - 1 = yln[(1 + e^x)/2]

Therefore the equation of curve must be y² - 1 = yln[(1 + e^x)/2]