Verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
$y = \sqrt{1 + x^{2}} : y^{'} = \frac{x y}{1 + x^{2}}$

Open in App

Solution

Let
$y = \sqrt{1 + x^{2}}$
Differentiating both sides w.r.t. $x$ we get
$y^{'} = (\sqrt{1 + x^{2}})^{'}$
$y^{'} = \frac{1}{2 \sqrt{1 + x^{2}}} \times 2 x$
$y^{'} = \frac{x}{\sqrt{1 + x^{2}}}$
Taking LHS
$y^{'} = \frac{x}{\sqrt{1 + x^{2}}}$
$y^{'} = \frac{1}{\sqrt{1 + x^{2}}} \times \frac{y}{y}$
(Multiplyig and dividing by $y$ )
Putting $y = \sqrt{1 + x^{2}}$ in denominator then we get,
$y^{'} = \frac{x y}{\sqrt{1 + x^{2}} \times \sqrt{1 + x^{2}}}$
$y^{'} = \frac{x y}{1 + x^{2}}$
Thus $L H S = R H S$
Hence verified.

Final answer:
Hence, the function $y = \sqrt{1 + x^{2}}$ is a solution of the differetial equation $y^{'} = \frac{x y}{1 + x^{2}}$

Suggest Corrections

Similar questions

y = x^{2} + 2 x + C : y^{'} - 2 x - 2 = 0

y = \sqrt{a^{2} - x^{2}} x \in (- a, a) : x + y \frac{d y}{d x} = 0 (y \neq 0)

y = e^{x} + 1 : y " - y^{'} = 0

y - cos y = x : (y sin y + cos y + x) y^{'} = y

y = x sin x : x y^{'} = y + x \sqrt{x^{2} - y^{2}} (x \neq 0 and x > y or x < - y)

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program