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

Differential Equation

Q.2. Solve the following differential equation :

dy/dx = e^x+yx+y + x².ey.

Solution :

We have dy/dx = e^x+y + x².e^y = e^x .ey + x².e^y = (ex + x²).e^y
Seperating the variable, we get
∫e–ydy = (ex + x²)dx + c
Or, – y^e–y = e^x + x³/3 + c => e^x + e^y + x³/3 + c = 0 [Ans.]

Q.3. Solve the following differential equation :

x(1 + y²)dx – y(1 + x²)dy = 0, given that y = 0 when x = 1.

Solution :

We have, x(1 + y²)dx – y(1 + x²)dy = 0 ------------- (1)
Or, [x/(1 + x²)] dx = [y/(1 + y²)] dy
Integrating, we get
∫[y/(1 + y²)] dy = ∫[x/(1 + x²)] dx
Or, 1/2log(1 + y²) = 1/2log(1 + x²) + c ------------- (2)
When x = 1, y = 0; putting in (2) we get
1/2log1 = 1/2log2 + c => 0 = 1/2log2 + c => c = – 1/2log2.
From (2) we get,
1/2log(1 + y²) = 1/2log(1 + x²) – 1/2log2
=> log(1 + y²) = log[(1 + x²)/2]
=> 2(1 + y²) = 1 + x²
=> x² – 2y² – 1 = 0 [Ans.]

Q.4. Solve the following differential equation : (1 + e²x)dy + (1 + y²)ex dx = 0.

Solution :

We have, (1 + e^2x)dy + (1 + y²)e^x dx = 0
Or, (1 + e^2x)dy = – (1 + y²)e^x dx
Or, dy/(1 + y²) = – ex/(1 + e^2x) dx
Integrating both sides, we get
∫dy/(1 + y²) = – ∫ex/(1 + e^2x) dx
[Put ex = t => ex dx = dt]
tan ^-1y = – ∫dt/(1 + t²)
tan ^-1y = – tan ^-1 t + c
tan^-1y + tan ^-1ex = c. [Ans.]

9.5. Homogeneous Differential Equations.

Q.1. Solve the following differential equation : (y² – x²) dy = 3xy dx.

Solution :

We have, (y² – x²) dy = 3xy dx
Or, dy/dx = 3xy/(y² – x²) ------------ (1)
This is a homogeneous differential equation,
We put y = vx , then dy/dx = v + x dv/dx --------------- (2)
From (1) and (2) we get,
v + x dv/dx = (3x.vx)/(v²x² – x²)
= 3v/(v² – 1)
Or, x dv/dx = 3v/(v² – 1) – v
= (4v – v³)/(v² – 1)
Or, [(v² – 1)/v(4 – v²)]dv = dx/x
Integrating we get,
∫(v² – 1)/[v(2 – v)(2 + v)] dv = ∫dx/x + c
Or, ∫[–1/4v + 3/8. 1/(2 – v) – 3/8.1/(2 + v)] dv = ∫dx/x + c
[Resolving into partial fractions]
Or, – 1/4 log v – 3/8 log |2 – v| – 3/8 log |2 + v| = log x + c
Or, – 1/4log(y/x) – 3/8 log |4 – y2/x²| = log x + c. [Ans.]

Paper By Mr. M.P.Keshari
Email Id : [email protected]
Ph No. : 09434150289