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 = {x(2y – x)}/{x(2y + x)}, if y = 1 when x = 1.

Solution :

We have, dy/dx = {x(2y – x)}/{x(2y + x)}
= (2y – x )/(2y + x)
This is a homogeneous differential equation. Putting y = vx, we get
dy/dx = v + x dv/dx.
Therefore, v + xdv/dx = (2vx – x)/(2vx + x) = (2v – 1)/(2v + 1)
Or, xdv/dx = (2v – 1)/(2v + 1) – v
= (2v – 1 – 2v² – v)/(2v + 1)
= (– 2v² + v – 1)/(2v + 1)
= – (2v² – v + 1)/(2v + 1)
Integrating, we get
∫[(2v + 1)/(2v² – v + 1)]dv = – ∫dx/x + c
Putting 2v² – v + 1 = t, we get (4v – 1)dv = dt, and
1/2∫[(4v + 2)/(2v² – v + 1)]dv = – log|x| + c
Or, 1/2∫[(4v – 1 + 3)/(2v² – v + 1)]dv = – log|x| + c
Or, 1/2∫[(4v – 1)/(2v² – v + 1)]dv + 3/2∫dv/(2v² – v + 1) = – log|x| + c
Or, 1/2∫dt/t + 3/2∫dv/[2(v² – v/2 + 1/2)] = – log|x| + c
Or, 1/2log|t| + 3/4∫dv/[(v – 1/4)² – 1/16 + 1/2] = – log|x| + c
Or, 1/2log|t| + 3/4dv/[(v – 1/4)² + 7/16] = – log|x| + c
Or, 1/2log|2v² – v + 1| + 3/4∫dv/[(v – 1/4)² +(√7/4)2] = – log|x| + c
Or, 1/2log|2v² – v + 1| + (3/4).1/(√7/4) tan –1[(v – 1/4)/(√7/4)] = – log|x| + c
Or, 1/2log|2v² – v + 1| + (3/√7)tan –1[(4v – 1)/√7] = – log|x| + c
Or, 1/2log|2y²/x² – y/x + 1| + (3/√7)tan –1[(4y/x – 1)/√7] = – log|x| + c
Or, 1/2log|(2y² – xy + x²2)/x²| + (3/√7)tan –1[(4y – x)/x√7] = – log|x| + c
Or, 1/2log|2y² – xy + x²| – 1/2log x² + (3/√7)tan –1[(4y – x)/x√7] – log|x| + c
Or, 1/2log|2y² – xy + x²| – log|x| + (3/√7)tan –1[(4y – x)/x√7] = – log|x| + c
Or, 1/2log|2y² – xy + x²| + (3/√7)tan –1[(4y – x)/x√7] = c
Given x = 1, y = 1; we get
1/2log|2 – 1 + 1| + (3/√7)tan –1[(4 – 1)/√7] = c
Or, 1/2log2 + (3/√7)tan –1(3/√7) = c
Hence, required solution is
1/2log|2y² – xy + x²| + (3/√7)tan –1[(4y – x)/x√7]
= 1/2log2 + (3/√7)tan –1(3/√7). [Ans.]

Q.6. Solve the following differential equation : x² dy/dx = y² + 2xy. Given that y = 1 when x = 1.

Solution :

We have, x² dy/dx = y² + 2xy
Or, dy/dx = (y² + 2xy)/x².
This is a homogeneous differential equation,
put y = vx, then dy/dx = v + x dv/dx
Or, v + x dv/dx = (v²x² + 2x.vx)/x²
= v² + 2v
Therefore, x dv/dx = v² + v
Or, dv/(v² + v) = dx/x
Therefore, ∫dv/v(v + 1) = ∫dx/x [1/v(v + 1) = A/v + B/(v + 1), A = 1, B = –1]
Or, ∫dv/v – ∫dv/(v + 1) = ∫dx/x
Or, log|v| – log|v + 1| = log|x| + log c
Or, log|v/(v + 1)| = log|cx|
Therefore, v/(v + 1) = cx
Or, (y/x)/{(y/x) + 1} = cx
Or, y/(x + y) = cx
When x = 1, y = 1, c = 1/2.
Therefore, y = x²/(2 – x). [Ans.]

Paper By Mr. M.P.Keshari
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