SATHEE: Differential Equations 1 Question 10

Differential Equations 1 Question 10

10.

The order of the differential equation whose general solution is given by $y = (c_{1} + c_{2}) \cos (x + c_{3}) - c_{4} e^{x + c_{5}}$ , where $c_{1}, c_{2}, c_{3}, c_{4}, c_{5}$ are arbitrary constants, is

(a) 5

(b) 4

(d) 2

(1998, 2M)

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Answer:

Correct Answer: 10. ( c)

Solution:

Given, $y = (c_{1} + c_{2}) \cos (x + c_{3}) - c_{4} e^{x + c_{5}}$ …….(i)

$\Rightarrow y = (c_{1} + c_{2}) \cos (x + c_{3}) - c_{4} e^{x} \cdot e^{c_{5}}$

Now, let $c_{1} + c_{2} = A, c_{3} = B, c_{4} e^{c_{5}} = c$

$\Rightarrow y = A \cos (x + B) - c e^{x}$ …….(ii)

On differentiating w.r.t. $x$ , we get

$\frac{d y}{d x} = - A \sin (A + B) - c e^{x}$ …….(iii)

Again, on differentiating w.r.t. $x$ , we get

$\frac{d^{2} y}{d x^{2}} = - A \cos (x + B) - c e^{x}$ …….(iv)

$\Rightarrow \frac{d^{2} y}{d x^{2}} = - y - 2 c e^{x}$ …….(v)

$\Rightarrow \frac{d^{2} y}{d x^{2}} + y = - 2 c e^{x}$

Again, on differentiating w.r.t. $x$ , we get

$\frac{d^{3} y}{d x^{3}} + \frac{d y}{d x} = - 2 c e^{x}$ …….(vi)

$\Rightarrow \frac{d^{3} y}{d x^{2}} + \frac{d y}{d x} = \frac{d^{2} y}{d x^{2}} + y$

[from Eq. (v)]

which is a differential equation of order 3 .

Differential Equations 1 Question 10

10.

Answer:

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Differential Equations 1 Question 10

10.

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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