COMPASSLearning

Talk to us Register Login

CBSE Class 12 · 2015 · Differential EquationsLong Answer

Solve the differential equation

(tan^{- 1} y - x) d y = (1 + y^{2}) d x

.

Answer: $x = (tan^{- 1} y - 1) + C e^{- t a n^{- 1} y}$

Solution

Rewrite as linear in

x

(with

y

independent):

\frac{d x}{d y} + \frac{1}{1 + y ^{2}} x = \frac{tan ^{- 1} y}{1 + y ^{2}}

I.F. = e^{\int \frac{d y}{1 + y ^{2}}} = e^{t a n^{- 1} y}

x e^{t a n^{- 1} y} = \int e^{t a n^{- 1} y} \frac{tan ^{- 1} y}{1 + y ^{2}} d y

Put

t = tan^{- 1} y

:

\int t e^{t} d t = e^{t} (t - 1)

, so

x e^{t a n^{- 1} y} = e^{t a n^{- 1} y} (tan^{- 1} y - 1) + C \Rightarrow x = (tan^{- 1} y - 1) + C e^{- t a n^{- 1} y}

CBSE Class 12 Board · 2015 · Subjective

Practise more: all CBSE board PYQs on Differential Equations →

More board questions on Differential Equations

(a) Find the particular solution of the differential equation
$\frac{d y}{d x} - 2 x y = 3 x^{2} e^{x^{2}}; y (0) = 5$

OR
(b) Solve the following differential equation:
$x^{2} d y + y (x + y) d x = 0$
CBSE 2024View solution →
If $\frac{d}{d x} f (x) = 2 x + \frac{3}{x}$ and $f (1) = 1$ , then $f (x)$ is:
CBSE 2023View solution →
Degree of the differential equation $sin x + cos (\frac{d y}{d x}) = y^{2}$ is:
CBSE 2023View solution →
The integrating factor of the differential equation $(1 - y^{2}) \frac{d x}{d y} + y x = a y, (- 1 < y < 1)$ is:
CBSE 2023View solution →
Find the general solution of the differential equation: $(x y - x^{2}) d y = y^{2} d x$
CBSE 2023View solution →
Find the general solution of the differential equation: $(x^{2} + 1) \frac{d y}{d x} + 2 x y = x^{2} + 4$
CBSE 2023View solution →