COMPASSLearning

Talk to us Register Login

CBSE Class 12 · 2010 · Differential EquationsLong Answer

Find the particular solution of the differential equation satisfying the given condition:

\frac{d y}{d x} = y tan x, given that y = 1 when x = 0.

Answer: $y = sec x$

Solution

Separate variables:

\frac{d y}{y} = tan x d x \Rightarrow lo g ∣ y ∣ = lo g ∣ sec x ∣ + C_{1}

Apply

y = 1

at

x = 0

:

lo g 1 = lo g sec 0 + C_{1} \Rightarrow C_{1} = 0

. Hence

y = sec x

CBSE Class 12 Board · 2010 · 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 →