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

Differential Equation

Q.5. Solve the following differential equation : dy/dx – y/x = 2x2.

Solution :

We have, dy/dx – y/x = 2x². -------------- (1)
This is a linear equation where, P = –1/x and Q = 2x².
I.F. = e^∫Pdx = e^{∫(–1/x)dx} = e^–logx = e^log(1/x) = 1/x.
Therefore, solution of (1) is given by
y. 1/x = ∫2x².1/x.dx + c
= ∫2xdx + c
= x² + c. [Ans.]

Q.6. Solve the differential equation : (1 – x²)dy/dx + xy = ax.

Solution :

The given differential equation is
(1 – x²)dy/dx + xy = ax
Or, dy/dx + x/(1 – x²).y = ax/(1 – x²) ---------- (1)
This is a linear differential equation, where, P = x/(1 – x²).
Therefore, Integrating factor = e∫Pdx
= e^{∫[x/(1 – x2)]dx}
= e ^{–1/2∫[–2x/(1 – x2)]dx}
= e ^{–1/2log(1 – x2)}
= e^log(1 – x²)–1/2
= 1/√(1 – x²).
Therefore, solution of (1) is
y.[1/√(1 – x²)] = ∫[{ax/(1 – x²)}{1/√(1 – x²)}] dx + c
= a∫[x/(1 – x²)3/2] dx + c = a∫(–1/2) dt/t^3/2 + c
[Put 1 – x² = t ]
= –a/2∫t –3/2dt + c
= –a/2[t –3/2/–1/2] + c
= a/√t + c
= a/√(1 – x²) + c
=> y = a + c√(1 – x²). [Ans.]

Q.7. Solve the differential equation : dy/dx + 2y tan x = sin x, given that y = 0 if x = π/3.

Solution :

We have, dy/dx + 2tan x.y = sin x -------------- (1)
This is a linear differential equation, where, P = 2 tan x.
Integrating factor = e∫Pdx
= e^{2∫tan x dx}
= e^{2logsec x}
= e^{logsec2 x}
= esec² x.
Therefore, the solution of (1) is
y.sec² x = ∫sin x sec² x dx + c
= ∫sec x tan x dx + c
= sec x + c
y = 0 if x = π/3 => cos x = 1/2
=> sec x = 2.
Therefore, 0.(2)² = 2 + c => c = –2.
=> ysec² x = sec x – 2
=> y = cos x – 2 cos² x. [Ans.]

Paper By Mr. M.P.Keshari
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