Obtain the differential equation whose solution is
$y = x sin (x + A),$ A being constant.

{(x y_{1} - y)}_{1}^{2} + x^{2} y^{2} = x^{4}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

{(x y_{1} - y)}_{1}^{2} - x^{2} y^{2} = x^{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{(x y_{1} - y)}_{1}^{2} + x^{2} y^{2} = x^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

{(x y_{1} - y)}_{1}^{2} - x^{2} y^{2} = x^{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A ${(x y_{1} - y)}_{1}^{2} + x^{2} y^{2} = x^{4}$
Given, $y = x sin (x + A)$

Differentiating, we get
$y_{1} = sin (x + A) + x cos (x + A)$
$∴ x y_{1} - y = x^{2} cos (x + A) . . (1)$
Also $x y = x^{2} sin (x + A) . . . (2)$
Squaring and adding (1) and (2)
${(x y_{1} - y)}_{1}^{2} + x^{2} y^{2} = x^{4} .1 = x^{4}$

Suggest Corrections

Similar questions

{cos}^{- 1} (\frac{y}{b}) = log {(\frac{x}{n})}^{n}

(Here y_{2} \equiv \frac{d^{2} y}{d x^{2}}, y_{1} \equiv \frac{d y}{d x})

If y = 3 cos (log x) + 4 sin (log x), show that $x^{2} y_{2} + x y_{1} + y = 0$
where $y_{1}$ & $y_{2}$ are the successive derivatives of y.

y = e^{m {sin}^{- 1} x}

(1 - x^{2}) y_{2} - x y_{1} - m^{2} y = 0

(x + \sqrt{x^{2} - 1}) m

(1 - x^{2}) y_{2} - x y_{1} + m^{2} y = 0

Join BYJU'S Learning Program