Mathematics Questions and Answers – Linear First Order Differential Equations – 1

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Linear First Order Differential Equations – 1”.

1. What is the differential equation whose solution represents the family c(y + c)2 = x3?
a) [2x/3 *(dy/dx) – y][x/3 *dy/dx]2 = x3
b) [x/3 *(dy/dx) – y][x/3 *dy/dx]2 = x3
c) [2x/3 *(dy/dx) – y][ 2x/3 *dy/dx]2 = x3
d) [x/3 *(dy/dx) – y][ 2x/3 *dy/dx]2 = x3
View Answer

Answer: c
Explanation: The given family is c(y + c)2 = x3
Differentiating once, we get
c[2(y + c)]dy/dx = 3x2
=> 2x3 (y + c)/(y + c)2 * dy/dx = 3x2
=> 2x3/(y + c) * dy/dx = 3x2
Or, 2x (y + c)/(y + c)2 * dy/dx = 3
=> 2x/3 *[dy/dx] = (y + c)
=> c = 2x/3 *[dy/dx] – y
Substituting c back into equation (1), we get
[2x/3 *(dy/dx) – y][ 2x/3 *dy/dx]2 = x3
which is the required differential equation
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2. What is the value of the solution of dy/dx = (6x + 9y – 7)/(2x + 3y – 6)?
a) -c/3
b) 3c/2
c) 2c/3
d) -2c/3
View Answer

Answer: a
Explanation: dy/dx = (6x + 9y – 7)/(2x + 3y – 6)
So, dy/dx = (3(2x + 3y) – 7)/(2x + 3x – 6) ……….(1)
Now, we put, 2x + 3y = z
Therefore, 2 + 3dy/dx = dz/dx [differentiating with respect to x]
Or, dy/dx = 1/3(dz/dx – 2)
Therefore, from (1) we get,
1/3(dz/dx – 2) = (3z – 7)/(z – 6)
Or, dz/dx = 2 + (3(3z – 7))/(z – 6)
= 11(z – 3)/(z – 6)
Or, (z – 6)/(z – 3) dz = 11 dx
Or, ∫(z – 6)/(z – 3) dz = ∫11 dx
Or, ∫(1 – 3/(z – 3)) dz = 11x + c
Or, z – log |z – 3| = 11x + c
Or, 2x + 3y – 11x – 3log|2x + 3y -3| = c
Or, 3y – 9x – 3log|2x + 3y – 3| = c
Or, 3x – y + log|2x + 3y – 3| = -c/3

3. What is the order of the differential equation of the family of circles with one diameter along the line y = x axis?
a) 3
b) 2
c) 1
d) 0
View Answer

Answer: b
Explanation: Centre of the circles can be taken as (a, a) and radius as r for some real numbers a and r.
Thus, the family is a two parameter family.
Hence, order of the corresponding differential equation is 2.
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4. What will be the general solution of differential equation for 2x – y dx + 2y – x dy = 0?
a) 22x + 22y = c
b) 2(22x + 22y) = c
c) 3(22x + 22y) = c
d) 4(22x + 22y) = c
View Answer

Answer: a
Explanation: We have, 2x – y dx + 2y – x dy = 0
Or, (2x/2y)dx + (2y/2x)dy = 0
Integrating both sides,
∫22xdx + ∫22ydy = k
Taking k = c/2log2 as, k, c and 2log2 are constant.
Or, 22x/2log2 + 22y/2log2 = c/2log2
Or, 22x + 22y = c

5. What will be the value of dy/dx = (x + 2y + 3)/(2x + 3y + 4)?
a) [(2 + √3)/2√3 * (log (√3(y + 2) – (x – 1))) + (2 – √3)/2√3 * (log (√3(y + 2) – (x – 1)))]
b) [(2 + √3)/2√3 * (log (√3(y + 2) + (x – 1))) – (2 – √3)/2√3 * (log (√3(y + 2) + (x – 1)))]
c) [(2 + √3)/2√3 * (log (√3(y + 2) – (x – 1))) – (2 – √3)/2√3 * (log (√3(y + 2) – (x – 1)))]
d) [(2 + √3)/2√3 * (log (√3(y + 2) + (x – 1))) – (2 – √3)/2√3 * (log (√3(y + 2) – (x – 1)))]
View Answer

Answer: c
Explanation: Put x = X + h, Y = Y + k,
We have, dY/dX = (X + 2Y +(h + 2k + 3))/ 2X + 3Y + (2h + 3k + 4)
So, (a – b)x = (a – b)
To determine h and k we set,
2h + 3k + 4 = 0 and h + 2k + 3 = 0
=> h = 1 and k = – 2
Therefore, dY/dX = (X + 2Y) / (2X + 3Y)
Putting Y = VX, we get,
V + X dV/dX = (1 + 2V)/(2 + 3V)
= (1 + 2V)/(3V2 – 1)*dV = -dX/X
=> [(2 + √3)/(2(√3V – 1)) – (2 – √3)/(2(√3V – 1))] dV = -dX/X
Simplifying it further, we get;
[(2 + √3)/2√3 * (log (√3Y – X)) – (2 – √3)/2√3 * (log (√3Y – X))]
Where, X = x – 1 and Y = y + 2
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6. What will be the value of the differential equation dy/dx = (x + y)2/(x + 2)(y – 2)?
a) keY/X where, X = x + 2 and Y = y – 2
b) ke2Y/X where, X = x + 2 and Y = y – 2
c) ke2Y/X where, X = x and Y = 2y
d) ke2X/Y where, X = x – 2 and Y = y + 2
View Answer

Answer: b
Explanation: Let, X = x + 2 and Y = y – 2
Then, dY/dX = (X + Y)2/XY
So, let, Y = vX
dY/dX = v + x dv/dX
=>v + Xdv/dX = (v + 1)2/v
=>v/1 + 2v dv = dX/X
=> (1 – 1/1 + 2v)dv = 2dx/x
=> v – ½ log (1 + 2v) = 2 log X + c
This means, X4(1 + 2Y/X)
= ke2Y/X where, X = x +2 and Y = y – 2

7. What will be the value of x(x – 1)dy/dx – y = x2(x – 1)2?
a) xy = (x + 1)(x3/3 + c)
b) depends on x
c) depends on y
d) xy = (x – 1)(x3/3 + c)
View Answer

Answer: d
Explanation: x(x – 1)dy/dx – y = x2(x – 1)2
or, dy/dx – 1/(x(x – 1))*y = x(x – 1) ……(1)
we have, ∫- 1/(x(x – 1)) dx = – ∫ [1/(x – 1) – 1/x] dx
= – [log(x – 1) – log x]
= log x – log(x – 1)
= log(x/(x – 1))
Thus, integrating factor = e∫- 1/(x(x – 1)) dx = x/(x – 1)
Thus, multiplying both sides of (1) by x/(x – 1), we get,
x/(x – 1)*dy/dx – 1/(x – 1)2 * y = x2
or, d/dx[x/(x – 1) * y] = x2 …….(2)
integrating both sides of (2) we get,
x/(x – 1) * y = ∫x2dx = x3/3 + c
or xy = (x – 1)(x3/3 + c)
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8. What is the solution of the equation(y + x(√xy)(x + y))dx – (y + y(√xy)(x + y))dy = 0?
a) x2 + y2 = 2tan-1(√(y/x)) + c
b) x2 + y2 = 4tan-1(√(y/x)) + c
c) x2 + y2 = tan-1(√(y/x)) + c
d) x2 + y2 = 2tan-1(√(x/y)) + c
View Answer

Answer: b
Explanation: The given equation can be written as on simplifying,
xdx + ydy + (ydx – xdy)/(√xy)(x + y) = 0
the above equation can be written as,
or, 1/2d(x2 + y2) = (xdy – ydx)/(x2 (√y/x) (1 + y/x))
= 2/(1 + y/x)* d(√y/x)
Thus, x2 + y2 = 4tan-1(√(y/x)) + c

9. A normal is drawn at a point P(x, y) of a curve. It meets the x-axis at Q. If PQ is of constant
length k. Then what is the differential equation describing such curve?
a) y(dy/dx) = ± √(k2 + x2)
b) y(dy/dx) = ± √(k2 – 2y2)
c) y(dy/dx) = ± √(k2 – y2)
d) Depends on the value of k
View Answer

Answer: c
Explanation: Equation of the normal at a point P(x, y) is given by
Y-y = -1/(dy/dx)(X – x) ….(1)
Let the point Q at the x-axis be (x1 , 0).
From (1), we get
y(dy/dx) = x1 – x ….(2)
Now, giving that PQ2 = k2
Or, x1 – x + y2 = k2
=> y(dy/dx) = ± √(k2 – y2)
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10. A normal is drawn at a point P(x, y) of a curve. It meets the x-axis at Q. If PQ is of constant
length k. What is the equation of such curve passing through (0, k)?
a) x2 + 2y2 = k2
b) 2x2 + y2 = k2
c) x2 – y2 = k2
d) x2 + y2 = k2
View Answer

Answer: d
Explanation: Equation of the normal at a point P(x, y) is given by
Y-y = -1/(dy/dx)(X – x) ….(1)
Let the point Q at the x-axis be (x1 , 0).
From (1), we get
y(dy/dx) = x1 – x ….(2)
Now, giving that PQ2 = k2
Or, x1 – x + y2 = k2
=> y(dy/dx) = ± √(k2 – y2) ….(3)
(3) is the required differential equation for such curves,
Now solving (3) we get,
∫-dy/√(k2 – y2) = ∫-dx
Or, x2 + y2 = k2 passes through (0, k)

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He is Linux Kernel Developer & SAN Architect and is passionate about competency developments in these areas. He lives in Bangalore and delivers focused training sessions to IT professionals in Linux Kernel, Linux Debugging, Linux Device Drivers, Linux Networking, Linux Storage, Advanced C Programming, SAN Storage Technologies, SCSI Internals & Storage Protocols such as iSCSI & Fiber Channel. Stay connected with him @ LinkedIn | Youtube | Instagram | Facebook | Twitter