CBSE Guess > Papers > Important Questions > Class XII > 2010 > Maths > Mathematics By Mr. M.P.Keshari

Differential Equation

Q.8. Solve the following differential equation : cos² x dy/dx + y = tan x.

Solution :

We have the differential equation
cos² xdy/dx + y = tan x
Dividing both sides by cos² x, we get
dy/dx + y sec² x = sec² x tan x,
which is a linear differential equation of the form:
dy/dx + Py = Q.
Here P = sec² x, Q = sec² x tan x.
Integrating factor IF = e^∫Pdx = e^{∫sec2 x dx} = e^{tan x}
Hence, the required solution is y. IF = ∫Q. IF dx + c
Or, y.etan x = ∫sec² x. tan x etan x dx + c
Put tan x = t, then sec² x dx = dt
Or, y.e^{tan x} = ∫te^t d^t + c
= ∫te^t d^t – ∫[d/d^t(t)∫et dt] dt + c
= te^t∫1.e^t. d^t + c
= te^t – e^t + c
= e^t(t – 1) + c
= e^{tan x}(tan x – 1) + c
Therefore, y = tan x – 1 + ce^{-tan x} . [Ans.]

Q.9. Solve the following differential equation : (x² + 1) dy/dx + 2xy = √(x² + 4).

Solution :

We have, (x² + 1) dy/dx + 2xy = √(x² + 4)
Or, dy/dx + {2x/(x² + 1)}y = {√(x² + 4)}/(x² + 1)
This is a linear differential equation of the form :
dy/dx + Py = Q.
Here, P = 2x/(x² + 1), Q = {√(x² + 4)}/(x² + 1).
Integrating factor, I F = e^∫Pdx
= e^{∫[2x/(x2 + 1)]dx}
Put x² + 1 = t => 2xdx = dt
I F = e^∫dt/t = e^logt = t = x² + 1.
Therefore, required solution is y. I F = ∫Q. I F dx + c
Or, y.(x² + 1) = ∫[{√(x² + 4)}/(x² + 1)].(x² + 1).dx + c
Or, y.(x² + 1) = ∫[√(x² + 4)]dx
[Using, ∫[√(x² + a2)]dx = x/2√(x² + a²) + a²/2log|x + √(x² + a²)| + c]
= x/2√(x² + 4) + 4/2log|x + √(x² + 4)| + c
= x/2√(x² + 4) + 2log|x + √(x² + 4)| + c. [Ans.

Paper By Mr. M.P.Keshari
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