Homogeneous Linear Differential Equations (General Form of LDE)
Trending Questions
Q.
Explain the procedure to find [second order derivative] of any function y=f(x).
Q. The general solution of the differential equation (D2−4D+4)y=0 is of the form (givenD=ddxandC1, C2areconstants)
- C1e2x
- C1e2x+C2e−2x
- C1e2x+C2e2x
- C1e2x+C2xe2x
Q. The solution to the ordinary differential equation d2ydx2+dydx−6y=0 is
- y=C1e3x+C2e−2x
- y=C1e3x+C2e2x
- y=C1e−3x+C2e2x
- y=C1e−3x+C2e−2x
Q.
If is the solution of the differential equation, , then the maximum value of the function over is equal to
Q. The solution for the following differential equation with boundary conditions y(0)=2 and y′(1)=−3 is, d2ydx2=3x−2
- y=x33−x22+3x−6
- y=3x3−x22−5x+2
- y=x32−x2−5x2+2
- y=x3−x22+5x+32
Q. The solution of the differential equation d2ydt2+2dydt+y=0 with y(0)=y′(0)=1 is
- (2−t)et
- (1+2t)e−t
- (2+t)e−t
- (1−2t)et
Q. y=e−2x is a solution of the differential equation y′′+y′−2y=0
- True
- False
Q.
If and are non-zero complex numbers such that . Then, the value of .
Q. If d2ydt2+y=0 under the conditions y=1, dydt=0, when t=0 then y is equal to
- sint
- cost
- tant
- cott
Q. The complete solution of the ordinary differential equation d2ydx2+pdydx+qy=0 is y=c1e−x+c2e−3x
Then, p and q are
Then, p and q are
- p=3, q=3
- p=3, q=4
- p=4, q=3
- p=4, q=4
Q.
If and , then at is
Q. The particular solution of the initial value problem given below is d2ydx2+12dydx+36y=0 with y(0)=3 and dydx∣∣∣x=0=−36
- (3−18x)e−6x
- (3+25x)e−6x
- (3+20x)e−6x
- (3−12x)e−6x
Q. The differential equation d2ydx2+16y=0 for y(x) with the two boundary conditions dydx∣∣∣x=0=1 and dydx∣∣∣x=12=−1 has
- no solution
- exactly two solutions
- exactly one solution
- infinitely many solutions
Q. The solution for the differential equation d2xdt2=−9x with initial conditions x(0)=1 and dxdt∣∣∣t=0=1, is
- t2+t+1
- sin3t+13cos3t+23
- 13sin3t+cos3t
- cos3t+t
Q. Consider the differential equation d2x(t)dt2+3dx(t)dt+2x(t)=0
Given x(0)=20 and x(1)=10/e, where e=2.718, the value of x(2) is
Given x(0)=20 and x(1)=10/e, where e=2.718, the value of x(2) is
Q.
The solution of the differential equation is
Q. If H(x, y) is a homogeneous function of degree n, then x∂H∂x+y∂H∂y=nH.
- True
- False
Q. For the differential equation
d2xdt2+6dxdt+8x=0 with initial conditions x(0)=1 and dxdt∣∣∣t=0=0), the solution is
d2xdt2+6dxdt+8x=0 with initial conditions x(0)=1 and dxdt∣∣∣t=0=0), the solution is
- x(t)=2e−6t−e−2t
- x(t)=2e−2t−e−4t
- x(t)=−e−6t+2e−4t
- a(t)=e−2t+2e−4t
Q. The solution of d2ydx2+2dydx+17y=0; y(0)=1, dydx(π4)=0 in the range 0<x<π4 is given by
- e−x(cos4x+14sin4x)
- ex(cos4x−14sin4x)
- e−4x(cos4x−14sin4x)
- e−4x(cos4x−14sin4x)
Q. The maximum value of the solution y(t) of the differential equation y(t)+..y(t)=0 with the initial conditions .y(0)=1 and y(0)=1, for t≥0 is
- 1
- 2
- π
- √2
Q. If f(x, y, z) = (x2+y2+z2)12 then ∂2f∂x2+∂2f∂y2+∂2f∂z2 is equal to
- zero
- 1
- 2
- -3(x2+y2+z2)52
Q.
How do you read
Q. Find the solution of d2ydx2=y which passes through origin and point (ln2, 34)
- y=12ex−e−x
- y=12(ex+e−x)
- y=12(ex−e−x)
- y=12ex+e−x
Q. The differential equation, dydx+Py=Q, is a linear equation of first order only if,
- P is a constant but Q is a function of y
- P and Q are functions of y or constants
- P is a function of y but Q is a constant
- P and Q are functions of x or constants
Q. For the equation x′′(t)+3x′(t)+2x(t)=5, the solution x(t) approaches to the following values as t→∞
- 0
- 5/2
- 5
- 10
Q. The solution to the differential equation d2udx2−kdudx=0 where k is constant, subjected to the boundary conditions u(0)=0 and u(L)=U, is
- u=UxL
- u=U(1−ekx1−ekL)
- u=U(1−e−kx1−e−kL)
- u=U(1+ekx1+ekL)
Q. It is given that y′′+2y′+y=0, y(0)=0, y(1)=0. What is y(0.5)?
- 0
- 0.37
- 0.62
- 1.13
Q. Which ONE of the following is a linear non- homogenous differential equation, where x and y are the independent and dependent variables respectively?
- dydx+xy=e−x
- dydx+xy=0
- dydx+xy=e−y
- dydx+e−y=0
Q.
If , where and are constant, then is
a constant
a function of
a function of
a function of and both
Q. If the characteristic equation of the differential equation d2ydx2+2αdydx+y=0 has two equal roots, then the values of α are
- ±1
- 0, 0
- ±j
- ±1/2