Question

y =√1 + x²):y' = ((xy)/(1 + x²))

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Answer 1

Answer

From the question,it is given that y=$\sf\sqrt{1 + {x}^{2} }$

Differentiating both sides with respect to x,we get,

y' = $\sf{ \dfrac{d}{dx}}$ $\bigg( \sf\sqrt{1 + {x}^{2} } \bigg)$

$:\implies$ y' = $\sf{\dfrac{1}{ 2\sqrt{1 + {x}^{2} }}} \sf{ .\dfrac{d}{dx}}{(1 + {x}^{2}) }$

By differentiating (1 + x²) we get,

$:\implies$y' = $\sf\dfrac{2x}{ 2\sqrt{1 + {x}^{2} } }$

On simplifying we get,

$:\implies$ y' = $\sf\dfrac{x}{ \sqrt{1 + {x}^{2} } }$

By multiplying and dividing √(1 + x²)

$:\implies$ y' = $\sf{ \dfrac{x}{1 + {x}^{2} } \times \sqrt{1 + {x}}^{2} }$

Substituting the value of √(1 + x²)

$:\implies$ y' = $\sf{ \dfrac{x}{1 + {x}^{2}}.y}$

$:\implies$ y' = $\sf{ \dfrac{xy}{1 + {x}^{2} }}$

Therefore,LHS = RHS

Therefore,the given function is a solution of the given differential equation.

hope it helps uh :)

Answer 2

Question

y =√1 + x²):y' = ((xy)/(1 + x²))

Solution

y = √1 + x²

dy/dx = d(√1 + x²)/dx

= 1/2√1 + x² × 2x

= x/√1 + x²

Now,We have to Verify

y' = xy/1 + x²

Taking L.H.S

y' = x/1√1 + x²

= x/√1 + x² × y/z (Multiplying and dividing by y)

= xy/√1 + x² × √1 + x² (Using by y = √1 - x² in denominator)

= xy/1 + x²

= R.H.S

Hence‚Verified...!!!!