SATHEE: Differential Equations 3 Question 15

Differential Equations 3 Question 15

15.

Determine the equation of the curve passing through the origin in the form $y = f (x)$ , which satisfies the differential equation $\frac{d y}{d x} = \sin (10 x + 6 y)$

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Answer:

Correct Answer: 15. ( $\frac{1}{3} \tan^{- 1} [\frac{4}{5} \tan (4 x + \tan^{- 1} \frac{3}{4}) - \frac{3}{5}] - \frac{5 x}{3}$ )

Solution:

Given,

$\frac{d y}{d x} = \sin (10 x + 6 y)$

Let $10 x + 6 y + t$ …….(i)

$\Rightarrow 10 + 6 \frac{d y}{d x} = \frac{d t}{d x}$

$\Rightarrow \frac{d y}{d x} = \frac{1}{6} (\frac{d t}{d x} - 10)$

Now, the given differential equation becomes

$\sin t = \frac{1}{6} (\frac{d t}{d x} - 10)$

$\Rightarrow 6 \sin t = \frac{d t}{d x} - 10$

$\Rightarrow \frac{d t}{d x} = 6 \sin t + 10$

$\Rightarrow \frac{d t}{6 \sin t + 10} = d x$

On integrating both sides, we get

$\frac{1}{2} \int \frac{d t}{3 \sin t + 5} = x + c$ …….(ii)

Let $I_{1} = \int \frac{d t}{3 \sin t + 5} = \int \frac{d t}{3 (\frac{2 \tan t / 2}{1 + \tan^{2} t / 2}) + 5}$

$= \int \frac{(1 + \tan^{2} t / 2) d t}{(6 \tan \frac{t}{2} + 5 + 5 \tan^{2} \frac{t}{2})}$

Put $\tan t / 2 = u$

$\Rightarrow \frac{1}{2} \sec^{2} t / 2 d t = d u \Rightarrow d t = \frac{2 d u}{\sec^{2} t / 2}$

$\Rightarrow d t = \frac{2 d u}{1 + \tan^{2} t / 2} \Rightarrow d t = \frac{2 d u}{1 + u^{2}}$

$∴ I_{1} = \int \frac{2 (1 + u^{2}) d u}{(1 + u^{2}) (5 u^{2} + 6 u + 5)} = \frac{2}{5} \int \frac{d u}{u^{2} + \frac{6}{5} u + 1}$

$= \frac{2}{5} \int \frac{d u}{u^{2} + \frac{6}{5} u + \frac{9}{25} - \frac{9}{25} + 1}$

$= \frac{2}{5} \int \frac{d u}{u + {\frac{3}{5}}^{2} + {\frac{4}{5}}^{2}} = \frac{2}{5} \cdot \frac{5}{4} \tan^{- 1} \frac{u + 3 / 5}{4 / 5}$

$= \frac{1}{2} \tan^{- 1} \frac{5 u + 3}{4} = \frac{1}{2} \tan^{- 1} \frac{5 \tan t / 2 + 3}{4}$

On putting this in Eq. (ii), we get

$\begin{aligned} \frac{1}{4} \tan^{- 1} [\frac{5 \tan \frac{t}{2} + 3}{4}] = x + c \\ \Rightarrow & \tan^{- 1} [\frac{5 \tan \frac{t}{2} + 3}{4}] = 4 x + 4 c \\ \Rightarrow & \frac{1}{4} [5 \tan (5 x + 3 y) + 3] = \tan (4 x + 4 c) \\ \Rightarrow & 5 \tan (5 x + 3 y) + 3 = 4 \tan (4 x + 4 c) \end{aligned}$

When $x = 0, y = 0$ , we get

$5 \tan 0 + 3 = 4 \tan (4 c)$

$\begin{array}{ll} \Rightarrow & \frac{3}{4} = \tan 4 c \\ \Rightarrow & 4 c = \tan^{- 1} \frac{3}{4} \end{array}$

Then, $5 \tan (5 x + 3 y) + 3 = 4 \tan (4 x + \tan^{- 1} \frac{3}{4})$

$\Rightarrow \tan (5 x + 3 y) = \frac{4}{5} \tan (4 x + \tan \frac{3}{4}) - \frac{3}{5}$

$\Rightarrow 5 x + 3 y = \tan^{- 1} [\frac{4}{5} \tan (4 x + \tan^{- 1} \frac{3}{4}) - \frac{3}{5}]$

$\Rightarrow 3 y = \tan^{- 1} [\frac{4}{5} \tan (4 x + \tan^{- 1} \frac{3}{4}) - \frac{3}{5}] - 5 x$

$\Rightarrow y = \frac{1}{3} \tan^{- 1} [\frac{4}{5} \tan (4 x + \tan^{- 1} \frac{3}{4}) - \frac{3}{5}] - \frac{5 x}{3}$

Differential Equations 3 Question 15

15.

Answer:

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Differential Equations 3 Question 15

15.

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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