# Mathematics Questions and Answers – Linear First Order Differential Equations – 2

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This set of Mathematics Multiple Choice Questions for Engineering Entrance Exams focuses on “Linear First Order Differential Equations – 2”.

1. A curve passes through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve is in the first quadrant and has its area equal to 2. What is the differential equation?
a) dy/dx = [(xy + 2) ± √(1 + xy)]/ x2
b) dy/dx = [(xy – 2) ± √(1 + xy)]/ x2
c) dy/dx = [(xy – 2) ± √(1 – xy)]/ x2
d) dy/dx = [(xy + 2) ± √(1 – xy)]/ x2

Explanation: The equation of tangent to the curve y = f(x), at point (x, y), is
Y – y = dy/dx * (X – x) …..(1)
Where it meets the x axis, Y = 0 and X = (x – y/(dy/dx))
Where it meets the y axis, X = 0 and Y = (y – x/(dy/dx))
Also, the area of the triangle formed by (1) with the coordinate axes is 2, so that,
(x – y/(dy/dx))* (y – x/(dy/dx)) = 4
Or, (y – x/(dy/dx))2 – 4dy/dx = 0
Or, x2(dy/dx)2 – 2(xy – 2)dy/dx + y2 = 0
Solving for dy/dx we get,
dy/dx = [(xy – 2) ± √(1 – xy)]/ x2

2. A curve passes through (1, 1) such that the triangle formed by the coordinate axes and the tangent at any point of the curve is in the first quadrant and has its area equal to 2. What will be the equation of the curve?
a) xy = 2
b) xy = -1
c) x – y = 2
d) x + y = 2

Explanation: The equation of tangent to the curve y = f(x), at point (x, y), is
Y – y = dy/dx * (X – x) …..(1)
Where it meets the x axis, Y = 0 and X = (x – y/(dy/dx))
Where it meets the y axis, X = 0 and Y = (y – x/(dy/dx))
Also, the area of the triangle formed by (1) with the coordinate axes is 2, so that,
(x – y/(dy/dx))* (y – x/(dy/dx)) = 4
Or, (y – x/(dy/dx))2 – 4dy/dx = 0
Or, x2(dy/dx)2 – 2(xy – 2)dy/dx + y2 = 0
Solving for dy/dx we get,
dy/dx = [(xy – 2) ± √(1 – xy)]/ x2
Let, 1 – xy = t2
=> x(dy/dx) + y = -2t(dt/dx)
=> x2(dy/dx) = t2 – 1 – 2tx(dt/dx), so that (3) gives
t(x(dt/dx) – (t ± 1)) = 0
Hence, either t = 0
=>xy = 1 which is satisfied by (1, 1)
Or, x dt/dx = t ± 1
=> dx/x = dt/t ± 1
=> t ± 1 = cx
For x = 1, y = 1 and t = 0
=> c = ± 1, so the solution is
t = ± (x – 1) => t2 = (x – 1)2
Or, 1 – xy = x2 – 2x + 1
Or, x + y = 2
Thus, the two curves that satisfies are xy = 1 and x + y = 2

3. What will be the value of dy/dx – a/x * y = (x + 1)/x?
a) y = x/(1 – a) – 1/a + cxa
b) y = x/(1 + a) + 1/a + cxa
c) y = x/(1 – a) – 1/a – cxa
d) y = x/(1 + a) – 1/a + cxa

Explanation: dy/dx – a/x * y = (x + 1)/x …….(1)
Multiplying both sides of equation (1) by
e∫-a/xdx
= e-a log x
= elog x-a
= x-a
We get, x-ady/dx – x-a (a/x)y = x-a (x + 1)/x
Or, d/dx(y . x-a) = x-a + x-a – 1 …….(2)
Integrating both sides of (2) we get,
y. x-a = x-a + 1/(-a + 1) + x-a – 1 + 1/(-a -1 + 1) + c
= x-a.x/(1 – a) + x-a/-a + c
Or, y = x/(1 – a) – 1/a + cxa

4. What will be the differential equation form of √(a2 + x2)dy/dx + y = √(a2 + x2) – x?
a) a2 log (x + √(a2 – x2)) + c
b) a2 log (x + √( a2 + x2)) + c
c) a2 log (x – √( a2 + x2)) + c
d) a2 log (x – √( a2 – x2)) + c

Explanation: The given form of equation can be written as,
dy/dx + 1/√(a2 + x2) * y = (√(a2 + x2) – x)/√(a2 + x2) ……(1)
We have, ∫1/√(a2 + x2)dx = log(x + √(a2 + x2))
Therefore, integrating factor is,
e∫1/√(a2 + x2) = elog(x + √(a2 + x2))
= x + √(a2 + x2)
Therefore, multiplying both sides of (1) by x + √(a2 + x2) we get,
x + √(a2 + x2dy/dx + (x + √(a2 + x2))/ √(a2 + x2)*y = (x + √(a2 + x2))(√(a2 + x2) – x)/√(a2 + x2)
or, d/dx[x + √(a2 + x2)*y] = (a2 + x2) ………..(2)
Integrating both sides of (2) we get,
(x + √(a2 + x2) * y = a2∫dx/√(a2 + x2)
= a2 log (x + √(a2 + x2)) + c

5. What is the solution of dy/dx = (6x + 9y – 7)/(2x + 3y – 6)?
a) 3x – y + log|2x + 3y – 3| = -c/3
b) 3x – y + log|2x + 3y – 3| = c/3
c) 3x + y + log|2x + 3y – 3| = -c/3
d) 3x – y – log|2x + 3y – 3| = c/3

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

6. A particle starts from the origin with a velocity 5cm/sec and moves in a straight line, its acceleration at time t seconds being (3t2 – 5t)cm/sec2. What will be the velocity of the particle?
a) 27cm/sec
b) 28 cm/sec
c) 29 cm/sec
d) 30 cm/sec

Explanation: Let x cm be the distance of the moving particle from the origin and v cm/sec be its velocity at the end of t seconds. Then the acceleration of the particle at time t seconds = dv/dt and its velocity at that time = v = dx/dt.
By question, dv/dt = 3t2 – 5t
Or, dv = 3t2 dt – 5tdt
Or, ∫dv = 3∫t2 dt – 5∫t dt
Or, v = t3 – (5/2)t2 + c ……….(1)
Given, v = 5, when t = 0; hence putting these values in equation (1) we get, c = 5
Thus v = t3 – (5/2)t2 + 5
Or, dx/dt = t3 – (5/2)t2 + 5 ………..(2)
Thus, the velocity of the particle at the end of 4 seconds,
= [v]t = 4 = (43 – (5/2)42 + 5 ) cm/sec [putting t = 4 in (2)]
= 29 cm/sec

7. A particle starts from the origin with a velocity 5cm/sec and moves in a straight line, its acceleration at time t seconds being (3t2 – 5t)cm/sec2. What will be the distance from the origin at the end of 4 seconds?
a) 30(4/3)
b) 30(2/3)
c) 30
d) Unpredictable

Explanation: Let x cm be the distance of the moving particle from the origin and v cm/sec be its velocity at the end of t seconds. Then the acceleration of the particle at time t seconds = dv/dt and its velocity at that time = v = dx/dt.
By question, dv/dt = 3t2 – 5t
Or, dv = 3t2 dt – 5t dt
Or, ∫dv = 3∫t2 dt – 5∫t dt
Or, v = t3 – (5/2)t2 + c ……….(1)
Given, v = 5, when t = 0; hence putting these values in equation (1) we get, c = 5
Thus v = t3 – (5/2)t2 + 5
Or, dx/dt = t3 – (5/2)t2 + 5 ………..(2)
Or, dx = t3 dt – (5/2)t2 dt + 5 dt
Integrating this we get,
x = (1/4)t4 – (5/2)t3/3 + 5t + k ……….(3)
By the problem, x = 0, when t = 0; hence, from (3) we get, k = 0.
Thus, x = (1/4)t4 – (5/6)t3 + 5t ……….(4)
Thus, the velocity of the particle at the end of 4 seconds,
= [x]t = 4 = (1/4)44 – (5/6)43 + 5(4) [putting t = 4 in (4)]
= 30(2/3) cm

8. What is the solution of (y(dy/dx) + 2x)2 = (y2 + 2x2)[1 + (dy/dx)2]?
a) cx±1/√2 = y/x + √(y2 – 2x2)/x2
b) cx±√2 = y/x + √(y2 + 2x2)/x2
c) cx±1/2√2 = y/x + √(y2 – 2x2)/x2
d) cx±1/√2 = y/x + √(y2 + 2x2)/x2

Explanation: Here, y2(dy/dx)2 + 4x2 + 4xy(dy/dx) = (y2 + 2x2)[1 + (dy/dx)2]
=>dy/dx = y/x ± √(1/2(y/x)2) + 1
Let, y = vx
=> v + x dv/dx = v ± √(1/2(v)2) + 1
Integrating both sides,
±√dv/(√(1/2(v)2) + 1) = ∫dx/x
cx±1/√2 = y/x + √(y2 + 2x2)/x2

9. What is the equation of the curve passing through (1, 0) of (y(dy/dx) + 2x)2 = (y2 + 2x2)[1 + (dy/dx)2]?
a) √2x±1/√2 = y/x + √(y2 + 2x2)/x2
b) √2x±1/2√2 = y/x + √(y2 + 2x2)/x2
c) √2x√2 = y/x + √(y2 + 2x2)/x2
d) √2x = y/x + √(y2 + 2x2)/x2

Explanation: Here, y2(dy/dx)2 + 4x2 + 4xy(dy/dx) = (y2 + 2x2)[1 + (dy/dx)2]
=> dy/dx = y/x ± √(1/2(y/x)2) + 1
Let, y = vx
=> v + x dv/dx = v ± √(1/2(v)2) + 1
Integrating both sides,
±∫dv/(√(1/2(v)2) + 1) = ∫dx/x
cx±1/√2 = y/x + √(y2 + 2x2)/x2 (put v/√2 = tan t)
putting x = 1, y = 0, we get c = √2
So, the curve is given by,
√2x±1/√2 = y/x + √(y2 + 2x2)/x2

10. If, 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 kind of curve is passing through (0, k)?
a) Parabola
b) Hyperbola
c) Ellipse
d) Circle

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)
Thus, it is a circle.

Sanfoundry Global Education & Learning Series – Mathematics – Class 12.

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