The solution of the differential equation $e^{- x} (y + 1) d y + ({cos}^{2} x + sin 2 x) y d x = 0$ subjected to the condition that $y = 1$ when $x = 0$ is

y + log y + e^{x} {cos}^{2} x = 2

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

log (y + 1) + e^{x} {cos}^{2} x = 1

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

y + log y = e^{x} {cos}^{2} x

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

(y + 1) + e^{x} {cos}^{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 $y + log y + e^{x} {cos}^{2} x = 2$
We have,
$e^{- x} (y + 1) d y + ({cos}^{2} x - sin 2 x) y d x = 0$
$\Rightarrow \frac{(y + 1)}{y} d y = - e^{x} ({cos}^{2} x - sin 2 x) d x$
$\Rightarrow (1 + \frac{1}{y}) d y = - e^{x} ({cos}^{2} x - sin 2 x) d x$
On integrating both sides, we get
$y + log y = - e^{x} {cos}^{2} x - \int e^{x} sin 2 x d x + \int e^{x} sin 2 x d x + C$
$\Rightarrow y + log y = - e^{x} {cos}^{2} x + C$
Given, when $x = 0 \Rightarrow y = 1$
$\Rightarrow 1 + log 1 = - e^{0} \cdot {cos}^{2} 0 + C$
$\Rightarrow 1 + 0 = - 1 + C \Rightarrow C = 2$
Therefore, $y + log y + e^{x} {cos}^{2} x = 2$

Suggest Corrections

Similar questions

\int \frac{(x + 1) e^{x}}{\cos^{2} (x e^{x})} d x

(y + x + 5) d y = (y - x + 1) d x

y = e^{x} + 1

y^{''} - y = 0

y = x^{2} + 2 x + C

y^{'} - 2 x - 2 = 0

y = cos x + C

y^{'} + sin x = 0

y = \sqrt{1 + x^{2}}

y^{'} = \frac{x y}{1 + x^{2}}

y = A x

x y = y (x \neq 0)

y = x sin x

x y = y + x \sqrt{x^{2} y^{2}} (x \neq 0 a n d x > y o r x < y)

x y = log y + C

y^{'} = \frac{y^{2}}{1 - x y} (x y \neq 1)

y - cos y = x

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

x + y = {tan}^{- 1} y

y^{2} y^{'} + y^{2} + 1 = 0

y = \sqrt{a^{2} - x^{2}} x ϵ (- a, a)

x + y \frac{d y}{d x} = 0 (y \neq 0)

x \frac{d y}{d x} + y l o g y = x y e^{x}

log \frac{x + y}{2} = \frac{1}{2} (log x + log y)

x = y

Join BYJU'S Learning Program