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Which one of the following is correct if we differentiate the equation xy = ae^{x} + be^{x} two times?
We have xy = ae^{x} + be^{x} ……(1)
Differentiating (1) with respect to x, we get
x(dy/dx) + y = ae^{x} + be^{x} …..(2)
Differentiating (2) now, with respect to x, we get
x(d^{2}y/ dx) + dy/dx + dy/dx = ae^{x} + be^{x}
From (1),
ae^{x} + be^{x} = xy, so that we get
x(d^{2}y/ dx) + 2(dy/dx) = xy
which is the required differential equation.
What is the differential equation of all parabolas whose directrices are parallel to the xaxis?
The equation of family of parabolas is Ax^{2} + Bx + C = 0 where, A, B, C are arbitrary constant.
By differentiating the equation with respect to x till all the constants get eliminated,
Hence, d^{3}y/dx^{3} = 0
What will be the required solution of d^{2}y/dx^{2} – 3dy/dx + 4y = 0?
d^{2}y/dx^{2} – 3dy/dx + 4y = 0 …..(1)
Let, y = e^{mx} be a trial solution of (1), then,
⇒ dy/dx = me^{mx} and d^{2}y/dx^{2} = m^{2}e^{mx}
Clearly, y = e^{mx} will satisfy equation (1). Hence, we have,
m^{2}e^{mx} – 3m * e^{mx} – 4e^{mx} = 0
⇒ m^{2} – 3m – 4 = 0 (as e^{mx} ≠ 0) …….(2)
⇒ m^{2} – 4m + m – 4 = 0
⇒ m(m – 4) + 1(m – 4) = 0
Or, (m – 4)(m + 1) = 0
Thus, m = 4 or m = 1
Clearly, the roots of the auxiliary equation (2) are real and unequal.
Therefore, the required general solution of (1) is
y = Ae^{4x} + Be^{x} where A and B are constants.
What is the solution of the given equation (D + 1)^{2}y = 0 given y = 2 log_{e} 2 when x = log_{e} 2 and y = (4/3) log_{e}3 when x = log_{e}3?
(D + 1)^{2}y = 0
Or, (D^{2} + 2D+ 1)y = 0
⇒ d^{2}y/dx^{2} + 2dy/dx + y = 0 ……….(1)
Let y = e^{mx} be a trial solution of equation (1). Then,
⇒ dy/dx = me^{mx} and d^{2}y/dx^{2} = m^{2}e^{mx}
Clearly, y = e^{mx} will satisfy equation (1). Hence, we have
⇒ m^{2}.e^{mx} + 2m.e^{mx} + e^{mx} = 0
Or, m^{2} + 2m +1 = 0 (as, e^{mx} ≠ 0) ………..(2)
Or, (m + 1)^{2} = 0
⇒ m = 1, 1
So, the roots of the auxiliary equation (2) are real and equal. Therefore, the general solution of equation (1) is
y = (A + Bx)e^{x} where A and B are two independent arbitrary constants ……….(3)
Given, y = 2 log_{e} 2 when x = log_{e} 2
Therefore, from (3) we get,
2 log_{e} 2 = (A + B log_{e}2)e^{x}
Or, 1/2(A + B log_{e}2) = 2 log _{e}2
Or, A + B log_{e}2 = 4 log_{e}2 ……….(4)
Again y = (4/3) log_{e}3 when x = log_{e}3
So, from (3) we get,
4/3 log_{e}3 = (A + Blog_{e}3)
Or, A + Blog_{e}3 = 4log_{e}3 ……….(5)
Now, (5) – (4) gives,
B(log_{e}3 – log_{e}2) = 4(log_{e}3 – log_{e}2)
⇒ B = 4
Putting B = 4 in (4) we get, A = 0
Thus the required solution of (1) is y = 4xe^{x}
If, A and B are arbitrary constants then what will be the differential equation of y = Ax + B/x?
Given, y = Ax + B/x
⇒ xy = Ax^{2} + B ……….(1)
Differentiating (1) with respect to x, we get,
d(xy)/dx = d/dx(Ax^{2} + B)
or, xdy/dx + y = A * 2x ……….(2)
Differentiating again with respect to x, we get,
x*d^{2}y/dx^{2} + dy/dx + dy/dx = A*2 ……….(3)
Eliminating A from (2) and (3) we get,
x^{2} d^{2} y/dx^{2} + 2xdy/dx = 2Ax [multiplying (3) by x]
or, x^{2} d^{2} y/dx^{2} + 2xdy/dx = xdy/dx + y [using (2)]
or, x^{2} d^{2} y/dx^{2} + xdy/dx – y = 0
What will be the value of C if C the constant of the coefficient of the solution of the given equation (D + 1)^{2}y = 0 given y = 2 log_{e} 2 when x = log_{e} 2 and y = (4/3) log_{e}3 when x = log_{e}3?
(D + 1)^{2}y = 0
Or, (D^{2} + 2D+ 1)y = 0
⇒ d^{2}y/dx^{2} + 2dy/dx + y = 0 ……….(1)
Let y = e^{mx} be a trial solution of equation (1). Then,
⇒ dy/dx = me^{mx} and d^{2}y/dx^{2} = m^{2}e^{mx}
Clearly, y = e^{mx} will satisfy equation (1). Hence, we have
⇒ m^{2}.e^{mx} + 2m.e^{mx} + e^{mx} = 0
Or, m^{2} + 2m + 1 = 0 (as, e^{mx} ≠ 0) ………..(2)
Or, (m + 1)^{2} = 0
⇒ m = 1, 1
So, the roots of the auxiliary equation (2) are real and equal. Therefore, the general solution of equation (1) is
y = (A + Bx)e^{x} where A and B are two independent arbitrary constants ……….(3)
Given, y = 2 log_{e} 2 when x = log_{e} 2
Therefore, from (3) we get,
2 log_{e} 2 = (A + B log_{e}2)e^{x}
Or, 1/2(A + B log_{e}2) = 2 log _{e}2
Or, A + B log_{e}2 = 4 log_{e}2 ……….(4)
Again y = (4/3) log_{e}3 when x = log_{e}3
So, from (3) we get,
4/3 log_{e}3 = (A + Blog_{e}3)
Or, A + Blog_{e}3 = 4log_{e}3 ……….(5)
Now, (5) – (4) gives,
B(log_{e}3 – log_{e}2) = 4(log_{e}3 – log_{e}2)
⇒ B = 4
Putting B = 4 in (4) we get, A = 0
Thus the required solution of (1) is y = 4xe^{x}
So, C = 4
What is thedifferential equation whose solution represents the family y = ae^{3x} + be^{x}?
The original given equation is,
y = ae^{3x} + be^{x} ……….(1)
Differentiating the above equation, we get
dy/dx = 3ae^{3x} + be^{x} ……….(2)
d^{2}y/dx^{2} = 9ae^{3x} + be^{x} ……….(3)
Now, to obtain the final equation we have to make the RHS = 0,
So, to get RHS = 0, we multiply, (1) with 3 and (2) with 4,
Thus , 3y = 3ae^{3x} + 3be^{x} ……….(4)
And 4(dy/dx) = 12ae^{3x} – 4 be^{x} ……….(5)
Now, adding the above three equations,(i.e. (3) + (4) + (5)) we get,
d^{2}y/dx^{2} – 4dy/dx + 3y = 0
If y = t(x) be a differentiable function ᵾ x € R, then which of the following is always true?
(dy/dx)= (dx/dy)1
So, d^{2}y/dx^{2} = – (dx/dy)^{2} d/dx(dx/dy)
= – (dy/dx)^{2}(d^{2}x/dy^{2})(dy/dx)
⇒ d^{2}y/dx^{2} + (dy/dx)^{3} d^{2}x/dy^{2} = 0
What will be the general solution of the differential equation d^{2}y/dx^{2} = e^{2x}(12 cos3x – 5 sin3x)? (here, A and B are integration constant)
Given, d^{2}y/dx^{2} = e^{2x}(12 cos3x – 5 sin3x) ………(1)
Integrating (1) we get,
dy/dx = 12∫ e^{2x}cos3x dx – 5∫ e^{2x} sin3x dx
= 12 * (e^{2x}/2^{2} + 3^{2})[2cos3x + 3sin3x] – 5 * (e^{2x}/2^{2} + 3^{2})[2sin3x – 3cos3x] + A (A is integrationconstant)
So dy/dx = e^{2x}/13 [24 cos3x + 36 sin3x – 10 sin3x + 15 cos3x] + A
= e^{2x}/13(39 cos3x + 26 sin3x) + A
⇒ dy/dx = e^{2x}(3 cos3x + 2 sin3x) + A ……….(2)
Again integrating (2) we get,
y = 3*∫ e^{2x} cos3xdx + 2∫ e^{2x} sin3xdx + A ∫dx
y = 3*(e^{2x}/2^{2} + 3^{2})[2cos3x + 3sin3x] + 2*(e^{2x}/2^{2} + 3^{2})[2sin3x – 3cos3x] + Ax + B (B is integration constant)
y = e^{2x}/13(6 cos3x + 9 sin3x + 4 sin3x – 6 cos3x) + Ax + B
or, y = e^{2x} sin3x + Ax + B
Which of the following is the valid differential equation x = a cos(αt + β)?
Since, x = a cos(αt + β)
Therefore, dx/dt = a cos(αt + β)
And, d^{2}x/dt^{2} = a α^{2} cos(αt + β)
= α^{2} a cos(αt + β)
Or, d^{2}x/dt^{2} = α^{2} [as a cos(αt + β) = x]
So, d^{2}x/dt^{2} + α^{2}x = 0
If, A normal is drawn at a point P(x, y) of a curve. It meets the xaxis at Q. If PQ is of constant length k. What kind of curve is passing through (0, k)?
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 xaxis be (x_{1} , 0).
From (1), we get
y(dy/dx) = x_{1} – x ….(2)
Now, giving that PQ^{2} = k^{2}
Or, x_{1} – x + y^{2} = k^{2}
=>y(dy/dx) = ± √(k^{2} – y^{2}) ….(3)
(3) is the required differential equation for such curves,
Now solving (3) we get,
∫dy/√(k^{2} – y^{2}) = ∫dx
Or, x^{2} + y^{2} = k^{2} passes through (0, k)
Thus, it is a circle.
What is the solution of (y(dy/dx) + 2x)^{2} = (y^{2} + 2x^{2})[1 + (dy/dx)^{2}]?
Here, y^{2}(dy/dx)^{2} + 4x^{2} + 4xy(dy/dx) = (y^{2} + 2x^{2})[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 + √(y^{2} + 2x^{2})/x^{2}
A particle starts from the origin with a velocity 5cm/sec and moves in a straight line, its acceleration at time t seconds being (3t^{2} – 5t)cm/sec^{2}. What will be the velocity of the particle?
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 = 3t^{2} – 5t
Or, dv = 3t^{2} dt – 5tdt
Or, ∫dv = 3∫t^{2} dt – 5∫t dt
Or, v = t^{3} – (5/2)t^{2} + c ……….(1)
Given, v = 5, when t = 0; hence putting these values in equation (1) we get, c = 5
Thus v = t^{3} – (5/2)t^{2} + 5
Or, dx/dt = t^{3} – (5/2)t^{2} + 5 ………..(2)
Thus, the velocity of the particle at the end of 4 seconds,
= [v]_{t = 4} = (4^{3} – (5/2)4^{2} + 5 ) cm/sec [putting t = 4 in (2)]
= 29 cm/sec
What will be the differential equation form of √(a^{2} + x^{2})dy/dx + y = √(a^{2} + x^{2}) – x?
The given form of equation can be written as,
dy/dx + 1/√(a^{2} + x^{2}) * y = (√(a^{2} + x^{2}) – x)/√(a^{2} + x^{2}) ……(1)
We have, ∫1/√(a^{2} + x^{2})dx = log(x + √(a^{2} + x^{2}))
Therefore, integrating factor is,
e^{∫1/√(a2 + x2)} = e^{log(x + √(a2 + x2))}
= x + √(a^{2} + x^{2})
Therefore, multiplying both sides of (1) by x + √(a^{2} + x^{2}) we get,
x + √(a^{2} + x^{2}dy/dx + (x + √(a^{2} + x^{2}))/ √(a^{2} + x^{2})*y = (x + √(a^{2} + x^{2}))(√(a^{2} + x^{2}) – x)/√(a^{2} + x^{2})
or, d/dx[x + √(a^{2} + x^{2})*y] = (a^{2} + x^{2}) ………..(2)
Integrating both sides of (2) we get,
(x + √(a^{2} + x^{2}) * y = a^{2}∫dx/√(a^{2} + x^{2})
= a^{2} log (x + √(a^{2} + x^{2})) + c
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?
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, x^{2}(dy/dx)^{2} – 2(xy – 2)dy/dx + y^{2} = 0
Solving for dy/dx we get,
dy/dx = [(xy – 2) ± √(1 – xy)]/ x^{2}
Let, 1 – xy = t^{2}
⇒ x(dy/dx) + y = 2t(dt/dx)
⇒ x^{2}(dy/dx) = t^{2} – 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) => t^{2} = (x – 1)^{2}
Or, 1 – xy = x^{2} – 2x + 1
Or, x + y = 2
Thus, the two curves that satisfies are xy = 1 and x + y = 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 is the differential equation?
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, x^{2}(dy/dx)^{2} – 2(xy – 2)dy/dx + y^{2} = 0
Solving for dy/dx we get,
dy/dx = [(xy – 2) ± √(1 – xy)]/ x^{2}
What will be the value of dy/dx – a/x * y = (x + 1)/x?
dy/dx – a/x * y = (x + 1)/x …….(1)
Multiplying both sides of equation (1) by
e^{∫a/xdx}
= e^{a log x}
= e^{log xa}
= x^{a}
We get, x^{a}dy/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 + cx^{a}
What is the solution of dy/dx = (6x + 9y – 7)/(2x + 3y – 6)?
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 – 3log2x + 3y 3 = c
Or, 3y – 9x – 3log2x + 3y – 3 = c
Or, 3x – y + log2x + 3y – 3 = c/3
A particle starts from the origin with a velocity 5cm/sec and moves in a straight line, its acceleration at time t seconds being (3t^{2} – 5t)cm/sec^{2}. What will be the distance from the origin at the end of 4 seconds?
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 = 3t^{2} – 5t
Or, dv = 3t^{2} dt – 5t dt
Or, ∫dv = 3∫t^{2} dt – 5∫t dt
Or, v = t^{3} – (5/2)t^{2} + c ……….(1)
Given, v = 5, when t = 0; hence putting these values in equation (1) we get, c = 5
Thus v = t^{3} – (5/2)t^{2} + 5
Or, dx/dt = t^{3} – (5/2)t^{2} + 5 ………..(2)
Or, dx = t^{3} dt – (5/2)t^{2} dt + 5 dt
Integrating this we get,
x = (1/4)t^{4} – (5/2)t^{3}/3 + 5t + k ……….(3)
By the problem, x = 0, when t = 0; hence, from (3) we get, k = 0.
Thus, x = (1/4)t^{4} – (5/6)t^{3} + 5t ……….(4)
Thus, the velocity of the particle at the end of 4 seconds,
= [x]_{t = 4} = (1/4)4^{4} – (5/6)4^{3} + 5(4) [putting t = 4 in (4)]
= 30(2/3) cm
What is the equation of the curve passing through (1, 0) of (y(dy/dx) + 2x)^{2} = (y^{2} + 2x^{2})[1 + (dy/dx)^{2}]?
Here, y^{2}(dy/dx)^{2} + 4x^{2} + 4xy(dy/dx) = (y^{2} + 2x^{2})[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 + √(y^{2} + 2x^{2})/x^{2} (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 + √(y^{2} + 2x^{2})/x^{2}
Find the general solution of the differential equation (x, y≠3).
Given that,
Separating the variables, we get
log(y  3) = log(x  3) + logC_{1}
log(y  3)  log(x  3) = logC_{1}
= logC_{1}
= 0
y  3 = 0 is the general solution for the given differential equation.
Find the general solution of the differential equation .
Given that,
Separating the variables, we get
2 cosy dy = 3 sinx dx
Integrating both sides, we get
∫ 2 cosy dy = ∫ 3 sinx dx
2 siny = 3(cosx) + C
3 cosx + 2 siny = C
Find the general solution of the differential equation
Given that,
Separating the variables, we get
dy = (3e^{x }+ 2)dx
Integrating both sides, we get
(1)
y = 3e^{x }+ 2x + C which is the general solution of the given differential equation.
Find the particular solution of the differential equation given that y = 1/3 when x = 1.
Given that,
Separating the variables, we get
dy = (4x  2)^{2} dx
Integrating both sides, we get
Given that, y = 1/3 when x = 1
Therefore, equation (1) becomes,
Hence, the particular solution for the given differential solution is y =
Find the particular solution of the differential equation .
Separating the variables, we get
Integrating both sides, we get
First, for integrating logy/y
Let logy = t
Differentiating w.r.t y, we get
1/y dy=dt
∴
=
Hence, equation (1), becomes
Given y = 2, we get x = 2
Substituting the values in equation (2), we get
C = 0
Therefore, the equation becomes
∴ (logy)^{2}  (logx)^{2} = 0.
Find the particular solution for the differential equation given that, y = 1 when x = 1.
Given that,
Separating the variables, we get
7y dy = 3x^{2} dx
Integrating both sides, we get
Given that y = 1, when x = 1
Substituting the values in equation (1), we get
Hence, the particular solution of the given differential equation is:
⇒ 7y^{2 }= 2x^{3 }+ 5
Which of the following functions is the solution of the differential equation dydx + 2y = 0?
Consider the function y = e^{2x}
Differentiating both sides w.r.t x, we get
dy/dx = 2y
⇒
Which of the following functions is a solution for the differential equation xy’  y = 0?
Consider the function y = 2x
Differentiating w.r.t x, we get
y’= dy/dx = 2
Substituting in the equation xy’y, we get
xy’ y = x(2)  2x = 2x  2x = 0
Therefore, the function y = 2x is a solution for the differential equation xy’  y = 0.
Which of the following functions is a solution for the differential equation y” + 6y = 0?
Consider the function y = 5 cos3x
Differentiating w.r.t x, we get
y’ = dy/dx = 15 sin3x
Differentiating again w.r.t x, we get
y” = d^{2}y/dx^{2} = 30 cos3x
⇒ y” + 6y=0.
Hence, the function y = 5 cos3x is a solution for the differential equation y” + 6y = 0.
Which of the following differential equations given below has the solution y = logx?
Consider the function y = logx
Differentiating w.r.t x, we get
Differentiating (1) w.r.t x, we get
∴
=
The number of arbitrary constants in a particular solution of a fourth order differential equation is ______
The number of arbitrary constants for a particular solution of nth order differential equation is always zero.
Which of the following is not a possible ordered pair for a matrix with 6 elements.
The possible orders in which the matrix with 6 elements can be formed are 2×3, 3×2, 1×6, 6×1. Therefore, the possible orders pairs are (2,3), (3,2), (1,6), (6,1). Thus, (3,1) is not possible.
How many arbitrary constants will be there in the general solution of a second order differential equation?
The number of arbitrary constants in a general solution of a nth order differential equation is n.
Therefore, the number of arbitrary constants in the general solution of a second order D.E is 2.
Which of the following functions is a solution for the differential equation dy/dx 14x = 0?
Consider the function y = 7x^{2}
Differentiating w.r.t x, we get
dy/dx 14x
⇒ dy/dx 14x = 0
Hence, the function y = 7x^{2} is a solution for the differential equation dydx  14x = 0
Which of the following differential equations has the solution y = 3x^{2}?
Consider the function y = 3x^{2}
Differentiating w.r.t x, we get
Differentiating (1) w.r.t x, we get
∴
Hence, the function y = 3x^{2} is a solution for the differential equation = 0
The number of rows (m) and the number of columns (n) in the given matrix A= is 2. Therefore, the order of the matrix is 2×2(m×n)
What is the difference property of definite integrals?
The sum difference property of definite integrals is
What is the constant multiple property of definite integrals?
The constant multiple property of definite integrals is
The zerolength interval property is one of the properties used in definite integrals and they are always positive. The zerolength interval property is .
is a property of definite integrals. is called as adding intervals property used to combine a lower limit and upper limit of two different integrals.
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