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

Differential Equation

Q.2. Solve the following differential equation : 2xy dx + (x² + 2y²) dy = 0.

Solution :

The given differential equation is
2xy dx + (x² + 2y²) dy = 0
Or, dy/dx = –2xy/(x² + 2y²) --------------------- (1)
This is a linear homogeneous differential equation.
Put y = v x, then dy/dx = v + x.dv/dx ------------ (2)
From (1) and (2) we get
v + x.dv/dx = – 2x(vx)/(x² + 2v²x²)
= – 2v/(1 + 2v2)
Or, x dv/dx = – v – 2v/(1 + 2v²)
= – (3v + 2v³)/(1 + 2v²)
Or, [(1 + 2v²)/(3v+ 2v³)].dv + dx/x = 0
Integrating, we get
∫[(1 + 2v²)/(3v + 2v³)] dv + ∫dx/x = log k
Or, 1/3∫dt/t + log x = log k [Putting 3v + 2v³ = t]
Or, 1/3 log t + log x = log k
Or, t1/3.x = k
=> tx³ = k³ = c
Or, (3v + 2v³) x³ = c
Or, 3vx³ + 2v³x³ = c
Replacing v = y/x, we get
3x²y + 2y³ = c. [Ans.]

Q.3. Solve the following differential equation : x dy/dx – y + x tan(y/x) = 0.

Solution :

We have, the given differential equation is
x dy/dx – y + x tan(y/x) = 0
Or, dy/dx – (y/x) + tan (y/x) = 0
Put y = vx , then dy/dx = v + x dv/dx
Therefore, equation (1) reduces to
v + x dv/dx – v + tan v = 0
Or, x dv/dx + tan v = 0
Or, cot v dv + dx/x = 0
Or, ∫cot v dv + ∫dx/x = log c
Or, log sin v + log x = log c
=> x sin v = c => x sin(y/x) = c. [Ans.]

Q.4. Solve the following differential equation : (x² – y²)dx + 2 xydy = 0, given that y = 1 when x = 1.

Solution :

We have, (x² – y²)dx + 2 xydy = 0
Or, dy/dx = (y² – x²)/2xy
This is a homogeneous differential equation, put y = vx, then dy/dx = v + xdv/dx.
Therefore, the given equation reduces to
v + xdv/dx = (v²x² – x²)/2x.vx
= (v² – 1)/2v
Or, xdv/dx = (v² – 1)/2v – v
= (v² – 1 – 2v²)/2v
= (– v² – 1)/2v
Or, 2v/(v2 + 1)dv = – dx/x
Integrating both sides, we get
∫2v/(v² + 1)dv = – ∫dx/x
Put v² + 1 = t, then 2v dv = dt ∫dt/t = – log|x| + c
Or, log|t| = – log|x| + c
Or, log|v² + 1| = – log|x| + c
Or, log|y2/x² + 1| = – log|x| + c
Or, log|(y² + x²)/x²| = – log|x| + c
Or, log|x² + y²| – log|x²| = – log|x| + c
Or, log|x² + y²| – 2log|x| = – log|x| + c
Or, log|x² + y²| = log|x| + c ----------------------- (1)
Putting x = 1 and y = 1, we get
log|1 + 1| = log|1| + c => c = log 2.
Putting the value of c, log|x² + y²| = log|x| + log 2
Or, log|x² + y²| = log|2x|
Therefore, x² + y² = 2x. [Ans.]

Paper By Mr. M.P.Keshari
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