solution of non-homogeneous linear differential equations with constant coefficients using operator method, method of variation of parameters, method of undetermined coefficient, solution of Euler-Cauchy equation, simultaneous differential equations by operator method MCQs

1. Using the operator method, the solution to (D² + 3D + 2)y = e^x is:

a) y = c₁e^-x + c₂e^-2x + (1/6)e^x b) y = c₁e^x + c₂e^2x + (1/6)e^-x c) y = c₁cosx + c₂sinx + e^x d) y = c₁e^-x + c₂xe^-x + e^x

Correct Answer: a) y = c₁e^-x + c₂e^-2x + (1/6)e^x

Explanation:

The given differential equation is:

\( (D² + 3D + 2)y = e^x \)

1. First, find the complementary function (CF) by solving the characteristic equation:

\( D² + 3D + 2 = 0 \)

\( (D + 1)(D + 2) = 0 \)

The roots are \( D = -1, -2 \), so the CF is:

\( y_c = c_1e^{-x} + c_2e^{-2x} \)

2. Next, find the particular integral (PI):

\( \text{Particular Integral } = \frac{e^x}{D² + 3D + 2} \)

Since the operator \( D - 1 \) can be applied, substitute \( D = 1 \):

\( = \frac{e^x}{1² + 3 × 1 + 2} = \frac{e^x}{6} = \frac{1}{6}e^x \)

3. Combining both results, the general solution is:

\( y = c_1e^{-x} + c_2e^{-2x} + \frac{1}{6}e^x \)

Hence, the correct answer is: Option (a).

2. For y'' + 4y = 3cos(2x), the appropriate form for particular solution is:

a) y_p = Acos(2x) b) y_p = Axcos(2x) + Bxsin(2x) c) y_p = Acosx + Bsinx d) y_p = A + Bcos(2x)

Correct Answer: b) y_p = Axcos(2x) + Bxsin(2x)

Explanation:

The given differential equation is:

\( y'' + 4y = 3\cos(2x) \)

1. First, note that the particular solution \( y_p \) is chosen to match the form of the non-homogeneous term, which is \( 3\cos(2x) \).

2. The complementary function will involve solutions to the characteristic equation:

\( r² + 4 = 0 ⟹ r = ±2i \)

The general complementary function is:

\( y_c = C_1\cos(2x) + C_2\sin(2x) \)

3. Since the non-homogeneous term \( 3\cos(2x) \) has the same form as the complementary function, we multiply by \( x \) to avoid redundancy. This results in the form:

\( y_p = Ax\cos(2x) + Bx\sin(2x) \)

Hence, the correct answer is: Option (b).

3. In variation of parameters for y'' + p(x)y' + q(x)y = g(x), the particular solution is:

a) y_p = u₁y₁ + u₂y₂ b) y_p = y₁ + y₂ c) y_p = u₁ + u₂ d) y_p = y₁y₂

Correct Answer: a) y_p = u₁y₁ + u₂y₂

Explanation:

The given differential equation is:

\( y'' + p(x)y' + q(x)y = g(x) \)

1. In the method of variation of parameters, the particular solution is expressed as a linear combination of the complementary function's solutions:

\( y_1 \) and \( y_2 \) are the solutions of the corresponding homogeneous equation:

\( y'' + p(x)y' + q(x)y = 0 \)

2. Instead of using constants, we introduce functions \( u_1(x) \) and \( u_2(x) \) as parameters:

\( y_p = u_1(x)y_1(x) + u_2(x)y_2(x) \)

3. This approach provides flexibility in satisfying the non-homogeneous equation, making it possible to determine the appropriate particular solution.

Hence, the correct answer is: Option (a).

4. The general solution of x²y'' - 2xy' + 2y = 0 is:

a) y = c₁x + c₂x² b) y = c₁x + c₂x^-1 c) y = c₁e^x + c₂e^2x d) y = c₁cos(lnx) + c₂sin(lnx)

Correct Answer: a) y = c₁x + c₂x²

Explanation:

The given differential equation is:

\( x^2y'' - 2xy' + 2y = 0 \)

1. This is a **Cauchy-Euler** equation. The standard form for such an equation is:

\( x^2y'' + axy' + by = 0 \)

2. To solve it, we substitute \( y = x^m \). Then:

\( y' = mx^{m-1}, \quad y'' = m(m-1)x^{m-2} \)

3. Substituting into the given equation, we get the characteristic equation:

\( m(m-1) - 2m + 2 = 0 \)

\( m^2 - 3m + 2 = 0 \)

4. Solving the quadratic equation:

\( (m-1)(m-2) = 0 ⟹ m = 1, 2 \)

5. The general solution is given by:

\( y = c_1x + c_2x^2 \)

Hence, the correct answer is: Option (a).

5. To solve dx/dt + dy/dt = t and dx/dt - dy/dt = 1 using operator method, we first:

a) Differentiate both equations b) Add and subtract the equations c) Integrate both equations d) Multiply them together

Correct Answer: b) Add and subtract the equations

Explanation:

1. The given system of differential equations is:

\( \frac{dx}{dt} + \frac{dy}{dt} = t \)

\( \frac{dx}{dt} - \frac{dy}{dt} = 1 \)

2. To simplify and solve using the operator method, the best approach is to add and subtract these two equations:

**Adding the equations:**

\( (\frac{dx}{dt} + \frac{dy}{dt}) + (\frac{dx}{dt} - \frac{dy}{dt}) = t + 1 \)

\( 2\frac{dx}{dt} = t + 1 ⟹ \frac{dx}{dt} = \frac{t + 1}{2} \)

**Subtracting the equations:**

\( (\frac{dx}{dt} + \frac{dy}{dt}) - (\frac{dx}{dt} - \frac{dy}{dt}) = t - 1 \)

\( 2\frac{dy}{dt} = t - 1 ⟹ \frac{dy}{dt} = \frac{t - 1}{2} \)

3. After this simplification, the differential equations can be solved independently.

Hence, the correct answer is: Option (b).

6. For y'' - y = xe^x, the correct form for particular solution is:

a) y_p = (Ax+B)e^x b) y_p = Axe^x c) y_p = x(Ax+B)e^x d) y_p = A + Be^x

Correct Answer: c) y_p = x(Ax+B)e^x

Explanation:

1. The given differential equation is:

\( y'' - y = x e^x \)

2. First, we identify the form of the non-homogeneous term \( x e^x \). Since \( e^x \) is a solution of the corresponding homogeneous equation, we multiply the particular solution by an extra factor of \( x \) to avoid duplication.

3. Therefore, the correct form for the particular solution is:

\( y_p = x(Ax + B)e^x \)

4. This choice ensures that we can solve for the unknown coefficients \( A \) and \( B \) correctly using substitution into the given equation.

Hence, the correct answer is: Option (c).

7. In variation of parameters, u₁' is calculated as:

a) -∫(y₂g/W)dx b) ∫(y₂g/W)dx c) -∫(y₁g/W)dx d) ∫(y₁g/W)dx

Correct Answer: a) -∫(y₂g/W)dx

Explanation:

1. In the variation of parameters method, the particular solution is assumed to be:

\( y_p = u_1 y_1 + u_2 y_2 \)

where \( y_1 \) and \( y_2 \) are the solutions of the corresponding homogeneous equation.

2. The formulas for \( u_1' \) and \( u_2' \) are given by:

\( u_1' = - \frac{y_2 g}{W} \) and \( u_2' = \frac{y_1 g}{W} \)

where:

\( g(x) \) is the non-homogeneous function.
\( W \) is the Wronskian of \( y_1 \) and \( y_2 \).

3. Therefore, the correct choice for \( u_1' \) is:

\( u_1' = -\frac{y_2 g}{W} \)

Hence, the correct answer is: Option (a).

8. The substitution used to convert Euler-Cauchy equation to constant coefficient form is:

a) x = e^t b) t = lnx c) Either a or b d) x = t²

Correct Answer: c) Either a or b

Explanation:

1. An **Euler-Cauchy equation** is a differential equation of the form:

\( x^2 y'' + a x y' + b y = 0 \)

2. To reduce it to a constant coefficient form, the standard substitution is:

\( x = e^t \) ⟹ \( t = \ln x \)

3. Using this substitution:

\( \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = \frac{dy}{dt} \cdot \frac{1}{x} \)

\( \frac{d^2y}{dx^2} = \frac{1}{x^2} \left( \frac{d^2y}{dt^2} - \frac{dy}{dt} \right) \)

4. This transformation converts the equation into one with constant coefficients.

5. Both \( x = e^t \) and \( t = \ln x \) are equivalent forms, so:

Hence, the correct answer is: Option (c).

9. For the system Dx + (D+1)y = e^t and (D-1)x + Dy = 0, the operator method involves:

a) Solving for x and y separately b) Eliminating one variable using determinants c) Integrating both equations d) Differentiating both equations

Correct Answer: b) Eliminating one variable using determinants

Explanation:

1. In the operator method, we treat \( D \) as the differential operator \( D = \frac{d}{dt} \).

2. The key idea is to eliminate one variable by manipulating the given system of equations using algebraic operations.

3. To achieve this, we can apply the following steps:

Multiply or rearrange the equations to form a system where one variable can be eliminated.
Use determinants (similar to the matrix form) to solve for the unknown functions.
Form a single differential equation in one variable and solve it.

4. This method is efficient because it avoids directly solving the coupled system using other complex methods.

Hence, the correct answer is: Option (b).

10. The method of undetermined coefficients cannot be used for:

a) e^x on right side b) sinx on right side c) lnx on right side d) polynomial on right side

Correct Answer: c) lnx on right side

Explanation:

1. The **method of undetermined coefficients** is a technique used to find particular solutions of non-homogeneous linear differential equations.

2. It works best when the right-hand side of the differential equation is a function that can be expressed as a sum of:

Exponentials like \( e^x \)
Polynomials
Trigonometric functions like \( \sin x \) and \( \cos x \)

3. However, functions like \( \ln x \) (logarithmic functions) are not suitable for the method of undetermined coefficients because their derivatives produce complex forms that are difficult to express using this method.

4. In such cases, the method of **variation of parameters** is more appropriate.

Hence, the correct answer is: Option (c).

UNIT 3 PART 1s

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