A first-order linear differential equation is of the form:
A
dy/dx + P(x)y = Q(x)
B
dy/dx = f(y/x)
C
d²y/dx² + P(x) dy/dx = 0
D
dy/dx + y² = 0
Analysis & Theory
A linear differential equation of first order has the form dy/dx + P(x)y = Q(x).
The standard method to solve dy/dx + P(x)y = Q(x) is:
A
Substitution y = vx
B
Integrating factor method
C
Separation of variables
D
Partial fraction method
Analysis & Theory
We solve linear differential equations using the integrating factor method.
The integrating factor (IF) for dy/dx + P(x)y = Q(x) is:
A
e^(∫Q dx)
B
e^(∫P dx)
C
∫P(x) dx
D
1/P(x)
Analysis & Theory
Integrating factor is IF = e^(∫P(x) dx).
After multiplying by the integrating factor, the left-hand side becomes:
A
d/dx [y * IF]
B
dy/dx only
C
Q(x) * IF
D
y² * IF
Analysis & Theory
Multiplying by IF makes the LHS an exact derivative: d/dx [y * IF].
General solution of dy/dx + P(x)y = Q(x) is:
A
y * IF = ∫Q(x) dx + C
B
y * IF = ∫Q(x) * IF dx + C
C
y = Q(x)/P(x)
D
y = Ce^(∫Q dx)
Analysis & Theory
The solution is y * IF = ∫Q(x) * IF dx + C.
If Q(x) = 0, the linear differential equation becomes:
A
Homogeneous linear equation
B
Non-homogeneous linear equation
C
Separable equation
D
Exact equation
Analysis & Theory
If Q(x)=0, it is a homogeneous linear equation.
Solution of dy/dx + P(x)y = 0 is:
A
y = Ce^(∫Q dx)
B
y = Ce^(∫P dx)
C
y = Ce^(-∫P dx)
D
y = Q(x)/P(x)
Analysis & Theory
For Q(x)=0, solution is y = Ce^(-∫P dx).
The equation dy/dx + 3y = 6 is:
A
Homogeneous linear
B
Non-homogeneous linear
C
Separable
D
Non-linear
Analysis & Theory
Because Q(x) ≠ 0, it is a non-homogeneous linear equation.
The solution of dy/dx + y = e^x is:
A
y = Ce^x + e^x
B
y = Ce^(-x) + e^(2x)/2
C
y = Ce^(-x) + e^x/2
D
y = Ce^x + e^(2x)
Analysis & Theory
Using IF = e^∫1 dx = e^x, solution is y*e^x = ∫e^(2x) dx + C ⇒ y = Ce^(-x) + e^(2x)/2.
Which of the following is NOT a linear differential equation?
A
dy/dx + y = x
B
dy/dx + y² = x
C
dy/dx + sin(x)y = 0
D
dy/dx + (1/x)y = ln(x)
Analysis & Theory
dy/dx + y² = x is non-linear because y² appears.