Important Questions

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Differential Equation

9.6.Linear Differential Equations.

Q.1. Solve the following differential equation : sin x dy/dx + cos x.y = cos x.sin2 x

Solution :

We have, sin x dy/dx + cos x.y = cos x.sin2 x
Or, dy/dx + cot x = cos x.sin x
This is a linear differential equation [dy/dx + P.y = Q]
Here P = cot x, Q = sin x.cos x
I.F. = e∫Pdx = e∫cot xdx = elog sin x = sin x
Therefore, the required solution of differential equation is
y. sin x = ∫cos x. sin x . sin x dx + c
= ∫ sin2 x cos x dx + c [Put sin x = t , then cos x dx = dt]
= ∫t2 dt + c
= t3/3 + c
= sin3 x/3 + c
Or, y = (1/3) sin2 x + c cosec x. [Ans.]

Q.2. Solve the following differential equation : dy/dx – y/x = 2x2.

Solution :

We have, dy/dx – (1/x).y = 2x2
This is a linear equation where, P = –1/x and Q = 2x2.
Therefore, I.F. = e∫P.dx = e∫–1/x.dx = e–log x = e log x –1 = x –1.
Hence, solution of the given differential equation is y.x –1 = ∫2x2.x –1 dx + c = 2∫x dx + c = 2.x2/2 + c = x2 + c.
Thus y = x3 + cx. [Ans.]

Q.3. Solve the following differential equation : dy/dx + (sec x).y = tan x.

Solution :

We have, dy/dx + (sec x) .y = tan x ------------------- (1)
This is a linear differential equation where, P = secx and Q = tan x.
Now I.F. = e∫Pdx
= e∫sec x dx
= elog(sec x + tan x)
= sec x + tan x.
Therefore, the solution of differential equation is :
y.(sec x + tan x) = ∫tan x (sec x + tan x)dx + c.
= ∫(sec x tan x + tan2 x)dx + c.
= ∫sec x tan x dx + ∫sec2 x dx – ∫1.dx + c.
= sec x + tan x – x + c.
Or, y = sec x + tan x – x + c. [Ans.]

Q.4. Solve the following differential equation : dy/dx + 2tan x .y = sin x.

Solution :

We have, dy/dx + 2tan x .y = sin x ----------------- (1)
This is a linear differential equation where, P = 2tan x and Q = sin x.
I.F. = e∫Pdx = e∫2tan x dx
= e2logsec x
= elogsec2 x
= sec2 x.
Therefore, solution of (1) is given by
y.sec2 x = ∫sin x.sec2 xdx + c
= ∫tan x.sec xdx + c
= sec x + c
Therefore, y = cos x + c cos2 x.

Maths Paper (With Solutions) By : Mr. M. P. Keshari
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Paper By Mr. M.P.Keshari
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