In this article, we focus exclusively on constant-coefficient ordinary differential equations, highlighting their significance.

In the previous article, we explored second-order homogeneous ODEs, providing the tools to solve complex problems. In this article, we'll shift our focus to nonhomogeneous equations, delving into them with detailed explanations, questions, and answers.

The general solution of a second-order non-homogeneous linear differential equation with constant coefficients can be expressed as:

y'' + a_1y' + a_2y = f(x)

y = y_c + y_p

where $y_p$ is any specific function that satisfies the non-homogeneous equation, and $y_c = C_1y_1 + C_2y_2$ is the general solution of the corresponding homogeneous equation: $y'' + a_1y' + a_2y = 0.$

Here, $y_c = C_1y_1 + C_2y_2$ is called the complementary solution (or homogeneous solution), and $y_p$ is called the particular solution (or non-homogeneous solution) of the equation

1. Method of Undetermined Coefficients

The particular solution is typically found through substitution, with the form depending on the nature of the differential equation. While these forms are effective in many cases, they may not always be sufficient.

Here are the most common scenarios:

- If $f(x) = k$ (a constant), we try $y_p = c$ .

- If $f(x) = kx$ , we try $y_p = ax + b$ .

- If $f(x) = kx^2$ , we try $y_p = ax^2 + bx + c$ , and so on for higher powers of $x$ .

- If $f(x) = k\sin x$ or $k\cos x$ , we try $y_p = a\sin x + b\cos x$ .

- If $f(x) = e^{kx}$ , we try $y_p = ce^{kx}$ . This method works provided that $k$ is not a root of the homogeneous equation’s characteristic equation. Otherwise, a modified approach is necessary.

1.1. Forcing Functions and Undetermined Coefficients

The method of undetermined coefficients works best when the forcing function (the non-homogeneous term on the right-hand side of a differential equation) belongs to a certain class of functions. These functions are typically of the following types:

1. Polynomials

The forcing function is of the form:

f(x) = a_n x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0

Examples:

- $f(x) = 3x^2 + 5x + 2$

- $f(x) = x^3 - 4$

2. Exponentials

The forcing function is of the form:

f(x) = A e^{\alpha x}

Examples:

- $f(x) = 5e^{2x}$

- $f(x) = e^{-x}$

3. Sines and Cosines

The forcing function is of the form:

f(x) = A \sin(\beta x) + B \cos(\beta x)

Examples:

- $f(x) = 3 \sin(2x)$

- $f(x) = 2 \cos(x)$

4. Combinations of Polynomials, Exponentials, and Trigonometric Functions

Forcing functions can also be combinations of the above types. For instance:

f(x) = (a_n x^n + \cdots + a_1 x + a_0) e^{\alpha x} \sin(\beta x)

Examples:

- $f(x) = x^2 e^{3x}$

- $f(x) = (3x - 1) \sin(4x)$

5. Products of Polynomials and Trigonometric Functions

Polynomials can also multiply sine and cosine terms:

f(x) = (a_n x^n + \cdots + a_1 x + a_0) \sin(\beta x) + (b_n x^n + \cdots + b_1 x + b_0) \cos(\beta x)

Examples:

- $f(x) = (x^2 + 3) \sin(2x)$

- $f(x) = x \cos(3x)$

1.2. Characteristic Equation and Roots

1. Form the Characteristic Equation:

For the homogeneous equation $y'' + a_1y' + a_2y = 0$ , the characteristic equation is $r^2 + a_1r + a_2 = 0$ .

2. Find the Roots:

Solve the characteristic equation to find the roots, which determine the nature of the complementary solution $y_c$ .

The reader can refer to the previous article to gain better insight on Linear non-homogeneous second-order ODEs(Constant Coefficients).

There are three possible scenarios based on the roots:

- Distinct real roots: If the roots $r_1$ and $r_2$ are real and distinct, the complementary solution is $y_c = C_1e^{r_1x} + C_2e^{r_2x}$ .

- Repeated real root: If the root is repeated, say $r_1 = r_2 = r$ , the complementary solution takes the form $y_c = (C_1 + C_2x)e^{rx}$ .

- Complex conjugate roots: If the roots are complex, say $r_1, r_2 = \alpha \pm \beta i$ , the complementary solution is $y_c = e^{\alpha x}(C_1\cos(\beta x) + C_2\sin(\beta x))$ .

1.3. Particular Solution and Forcing Functions

Once the characteristic roots are determined, the form of the particular solution $y_p(x)$ is influenced by the forcing function $f(x)$ . If $f(x)$ contains terms that correspond to solutions of the homogeneous equation, adjustments to the trial solution are required.

For example:

- If $f(x) = e^{kx}$ and $k$ is not a root of the characteristic equation, the trial solution $y_p = ce^{kx}$ works.

- If $f(x) = e^{kx}$ and $k$ is a root, we modify the trial solution to $y_p = x^n e^{kx}$ , where $n$ is the multiplicity of $k$ as a root.

1.4. Problem Solving: Method of Undetermined Coefficients

1. Solve the Complementary Equation:

- Begin by solving the corresponding homogeneous equation. This involves finding the roots of the characteristic equation and writing the general form of the complementary solution $y_c(x)$ .

2. Formulate the Particular Solution $y_p(x)$ :

- Analyze the form of the non-homogeneous term $f(x)$ and determine an appropriate form for the particular solution $y_p(x)$ .

3. Check for Overlap with the Complementary Solution:

- If any part of the proposed particular solution $y_p(x)$ appears in the complementary solution $y_c(x)$ , modify the form of $y_p(x)$ by multiplying by powers of $x$ .

4. Determine the Coefficients:

- Substitute the proposed form of $y_p(x)$ into the original differential equation. This results in a system of equations that you can solve to find the unknown coefficients.

5. Write the General Solution:

- Combine the complementary solution $y_c(x)$ and the particular solution $y_p(x)$ to obtain the complete general solution to the non-homogeneous equation.

Let's look at some examples:

2. Examples:

2.1. Example 1

\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + y = e^{2x}

Solution:

1. Solve the Homogeneous Equation:

y'' + 2y' + y = 0 \quad \Rightarrow \quad (m + 1)^2 = 0

Roots: $m = -1$ , repeated

y_c = (C_1 + C_2 x)e^{-x}

2. Particular Solution:

Since $f(x) = e^{2x}$ , $y_p = A e^{2x}$ .

Derivatives:

y_p' = 2A e^{2x}, \quad y_p'' = 4A e^{2x}

Substitute into the equation:

(4A e^{2x}) + 2(2A e^{2x}) + A e^{2x} = e^{2x}

Simplify:

y_p = \frac{1}{9}e^{2x}

3. General Solution:

y = y_c + y_p = (C_1 + C_2 x)e^{-x} + \frac{1}{9}e^{2x}

2.2. Example 2

\frac{d^2y}{dx^2} - y = x^2

Solution:

1. Solve the Homogeneous Equation:

y'' - y = 0 \quad \Rightarrow \quad m^2 - 1 = 0 \quad \Rightarrow \quad m = \pm 1

Homogeneous solution:

y_c = C_1 e^x + C_2 e^{-x}

2. Particular Solution:

Since $f(x) = x^2$ , $y_p = Ax^2 + Bx + C$ .

Derivatives:

y_p' = 2Ax + B, \quad y_p'' = 2A

Substitute into the equation:

2A - (Ax^2 + Bx + C) = x^2

Simplify:

(-A)x^2 + (-B)x + (2A - C) = x^2

Equate coefficients:

-A = 1, \quad -B = 0, \quad 2A - C = 0

Solve:

A = -1, \quad B = 0, \quad C = -2

y_p = -x^2 - 2

3. General Solution:

y = y_c + y_p = C_1 e^x + C_2 e^{-x} - x^2 - 2

Now we will see how to solve when the special case of overlap happens:

2.3. Example 3: Non-Homogeneous Equation with Multiplicity 1

\frac{d^2y}{dx^2} + y = \sin(x)

1. Solve the Homogeneous Equation:

y'' + y = 0

- Characteristic Equation:

m^2 + 1 = 0

- Roots:

m = \pm i

- General Solution of the Homogeneous Equation:

y_c = C_1 \cos x + C_2 \sin x

2. Formulate a Particular Solution $y_p(x)$ :

Since the non-homogeneous term $f(x) = \sin x$ is already part of the complementary solution, we directly adjust our particular solution to account for this overlap by multiplying by $x$ :

y_p = x(A \sin x + B \cos x)

3. Compute Derivatives:

- First derivative:

y_p' = A \sin x + B \cos x + x(A \cos x - B \sin x)

- Second derivative:

y_p'' = 2A \cos x - 2B \sin x - x(A \sin x + B \cos x)

4. Substitute into the Differential Equation:

Substituting $y_p$ and its derivatives into the original differential equation gives: and its derivatives into the original differential equation gives:

(2A \cos x - 2B \sin x - x(A \sin x + B \cos x)) + x(A \sin x + B \cos x) = \sin x

This simplifies to:

2A \cos x - 2B \sin x = \sin x

5. Equate Coefficients:

- Coefficient of $\cos x$ :

2A = 0 \quad \Rightarrow \quad A = 0

- Coefficient of $\sin x$ :

-2B = 1 \quad \Rightarrow \quad B = -\frac{1}{2}

6. Particular Solution:

Thus, the particular solution is:

y_p = -\frac{1}{2} x \cos x

7. Write the General Solution:

Combine the complementary solution and the particular solution:

y = y_c + y_p

The general solution is:

y = C_1 \cos x + C_2 \sin x - \frac{1}{2} x \cos x

2.4. Example 4: Non-Homogeneous Equation with Multiplicity 2

Differential Equation:

y'' + 4y' + 4y = e^{-2x}

First, we solve the associated homogeneous equation:

y'' + 4y' + 4y = 0

Characteristic Equation:

m^2 + 4m + 4 = 0 \quad \Rightarrow \quad (m + 2)^2 = 0

This gives a double root:

m = -2 \quad (\text{multiplicity } 2)

The complementary solution is:

y_c = (C_1 + C_2 x)e^{-2x}

Non-Homogeneous Term:

The non-homogeneous term is $e^{-2x}$ . Since this term is part of the complementary solution (with multiplicity 2), we need to multiply our guess for the particular solution by $x^2$ .

Particular Solution:

Thus, we assume a particular solution of the form.

y_p = Ax^2 e^{-2x}

Derivative:

y_p' = \frac{d}{dx} \left( A x^2 e^{-2x} \right) = A \left( 2x e^{-2x} - 2x^2 e^{-2x} \right) = A \left( 2x - 2x^2 \right) e^{-2x}

Second derivative $y_p''$ :

y_p'' = \frac{d}{dx} \left( A \left( 2x - 2x^2 \right) e^{-2x} \right)

\begin{align*} &= A \left( \left( 2 - 4x \right) e^{-2x} - 2 \left( 2x - 2x^2 \right) e^{-2x} \right) \\ &= A \left( 2 - 4x - 4x + 4x^2 \right) e^{-2x} \\ &= A \left( 2 - 8x + 4x^2 \right) e^{-2x} \end{align*}

Substitute $y_p$ , $y_p'$ , and $y_p''$ into the equation $y'' + 4y' + 4y = e^{-2x}$ .

y_p'' + 4y_p' + 4y_p = A \left( 2 - 8x + 4x^2 \right) e^{-2x} + 4A \left( 2x - 2x^2 \right) e^{-2x} + 4A x^2 e^{-2x}.

Simplifying:

= A \left( 2 - 8x + 4x^2 \right) e^{-2x} + A \left( 8x - 8x^2 \right) e^{-2x} + 4A x^2 e^{-2x}= A \left( 2 \right) e^{-2x}.

Thus, we have:

A \cdot 2 e^{-2x} = e^{-2x}.

Equating the coefficients of $e^{-2x}$ :

2A = 1 \quad \Rightarrow \quad A = \frac{1}{2}.

The general solution is the sum of the complementary and particular solutions:

y(x) = y_c + y_p = (C_1 + C_2 x)e^{-2x} + \frac{1}{2}x^2 e^{-2x}.

Note:

The non-homogeneous equation $y'' - 4y = x^2 e^{2x}$ has a complementary solution with multiplicity 1 for $e^{2x}$ .

The non-homogeneous equation $y'' + 4y' + 4y = e^{-2x}$ also has a complementary solution with multiplicity 2 for $e^{-2x}$ .

2.5. Example 5

\frac{d^2y}{dx^2} + 4y = 3\sin(2x)

1. Solve the Homogeneous Equation

\frac{d^2y}{dx^2} + 4y = 0

This is a second-order linear equation with constant coefficients. The characteristic equation is:

m^2 + 4 = 0

Solving for $m$

m = \pm 2i

This gives a solution for the homogeneous equation:

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

2. Find the Particular Solution

Now, for the non-homogeneous equation:

\frac{d^2y}{dx^2} + 4y = 3\sin(2x)

Since the forcing function $3 \sin(2x)$ is of the same form as the solution to the homogeneous equation, we multiply it by $x$ for the particular solution.

y_p = x(A \cos(2x) + B \sin(2x))

Now, find the first and second derivatives of $y_p$ :

y_p' = A \cos(2x) - 2Ax \sin(2x) + B \sin(2x) + 2Bx \cos(2x)

y_p'' = -4A \sin(2x) - 4Ax \cos(2x) + 4B \cos(2x) - 4Bx \sin(2x)

Substitute into the original equation:

Substitute $y_p$ and $y_p''$ into the original equation:

(-4A \sin(2x) - 4Ax \cos(2x) + 4B \cos(2x) - 4Bx \sin(2x)) + 4(x(A \cos(2x) + B \sin(2x))) = 3 \sin(2x)

Simplify the terms:

(-4A \sin(2x) + 4B \cos(2x)) = 3 \sin(2x)

Now equate the coefficients of $\sin(2x)$ and $\cos(2x)$ :

For $\sin(2x)$ :

-4A = 3 \quad \Rightarrow \quad A = -\frac{3}{4}

For $\cos(2x)$ :

4B = 0 \quad \Rightarrow \quad B = 0

Thus, the particular solution is:

y_p = -\frac{3}{4} x \cos(2x)

3. General Solution

Now, the general solution to the differential equation is the sum of the homogeneous and particular solutions:

y = y_c + y_p = C_1 \cos(2x) + C_2 \sin(2x) - \frac{3}{4} x \cos(2x)

2.6. Example 6

\frac{d^2y}{dx^2} + 6y = e^{-2x}

Solution:

1. Solve the Homogeneous Equation:

y'' + 6y = 0 \quad \Rightarrow \quad m^2 + 6 = 0 \quad \Rightarrow \quad m = \pm i\sqrt{6}

Complementary solution:

y_c = C_1 \cos(\sqrt{6}x) + C_2 \sin(\sqrt{6}x)

2. Particular Solution:

Since $f(x) = e^{-2x}$ , $y_p = Ae^{-2x}$ .

Derivatives:

y_p' = -2Ae^{-2x}, \quad y_p'' = 4Ae^{-2x}

Substitute into the equation:

4Ae^{-2x} + 6Ae^{-2x} = e^{-2x}

Simplify:

10A = 1 \quad \Rightarrow \quad A = \frac{1}{10}

y_p = \frac{1}{10}e^{-2x}

3. General Solution:

y = y_c + y_p = C_1 \cos(\sqrt{6}x) + C_2 \sin(\sqrt{6}x) + \frac{1}{10}e^{-2x}

2.7. Example 7

y'' - y = e^x

Solution:

1. Solve the Homogeneous Equation:

y'' - y = 0 \quad \Rightarrow \quad m^2 - 1 = 0 \quad \Rightarrow \quad m = \pm 1

Complementary solution:

y_c = C_1 e^x + C_2 e^{-x}

2. Particular Solution:

Since $f(x) = e^x$ , $y_p = Ax e^x$ .

Derivatives:

y_p' = A(e^x + xe^x), \quad y_p'' = A(2e^x + xe^x)

Substitute into the equation:

A(2e^x + xe^x) - A xe^x = e^x

Simplify:

A 2e^x = e^x \quad \Rightarrow \quad A = \frac{1}{2}

y_p = \frac{1}{2} xe^x

3. General Solution:

y = y_c + y_p = C_1 e^x + C_2 e^{-x} + \frac{1}{2} xe^x

2.8. Example 8

y'' + 9y = 3x

Solution:

1. Solve the Homogeneous Equation:

y'' + 9y = 0 \quad \Rightarrow \quad m^2 + 9 = 0 \quad \Rightarrow \quad m = \pm 3i

Complementary solution:

y_c = C_1 \cos(3x) + C_2 \sin(3x)

2. Particular Solution:

Since $f(x) = 3x$ , assume $y_p = Ax + B$ .

Derivatives:

y_p' = A, \quad y_p'' = 0

Substitute into the equation and equate coefficients:

9A = 3 \quad \Rightarrow \quad A = \frac{1}{3}, \quad 9B = 0 \quad \Rightarrow \quad B = 0

y_p = \frac{1}{3}x

3. General Solution:

y = y_c + y_p = C_1 \cos(3x) + C_2 \sin(3x) + \frac{1}{3}x

2.9. Example 9

y'' - 4y' + 3y = e^{x}

1. Solve the Homogeneous Equation

y'' - 4y' + 3y = 0

The characteristic equation is:

m^2 - 4m + 3 = 0

Factoring gives:

(m - 1)(m - 3) = 0 \quad \Rightarrow \quad m = 1, 3

Thus, the general solution of the homogeneous equation is:

y_c = C_1 e^x + C_2 e^{3x}

2. Particular Solution

Since $f(x) = e^{x}$ is part of the complementary solution, we adjust a particular solution of the form:

y_p = A x e^x

Calculating the derivatives:

y_p' = A e^x + A x e^x

y_p'' = A e^x + A e^x + A x e^x = 2A e^x + A x e^x

Substituting $y_p, y_p',$ and $y_p''$ into the original equation:

(2A e^x + A x e^x) - 4(A e^x + A x e^x) + 3(A x e^x) = e^x

Simplifying the left-hand side:

(2A e^x + A x e^x) - (4A e^x + 4A x e^x) + 3A x e^x

Combine like terms:

(2A - 4A)e^x + (A - 4A + 3A)x e^x = e^x

(-2A)e^x + 0 \cdot x e^x = e^x

Setting coefficients equal:

-2A = 1 \quad \Rightarrow \quad A = -\frac{1}{2}

Thus, the particular solution is:

y_p = -\frac{1}{2} x e^x

3. General Solution

y = y_c + y_p = C_1 e^x + C_2 e^{3x} - \frac{1}{2} x e^x

2.10. Example 10

y'' + y = e^{3x}

Solution:

1. Solve the Homogeneous Equation:

y'' + y = 0 \quad \Rightarrow \quad m^2 + 1 = 0 \quad \Rightarrow \quad m = \pm i

Complementary solution:

y_c = C_1 \cos x + C_2 \sin x

2. Particular Solution:

Since $f(x) = e^{3x}$ , $y_p = A e^{3x}$ .

Derivatives:

y_p' = 3A e^{3x}, \quad y_p'' = 9A e^{3x}

Substitute into the equation:

9A e^{3x} + A e^{3x} = e^{3x}

Simplify:

10A = 1 \quad \Rightarrow \quad A = \frac{1}{10}

y_p = \frac{1}{10}e^{3x}

3. General Solution:

y = y_c + y_p = C_1 \cos x + C_2 \sin x + \frac{1}{10}e^{3x}

2.11. Example 11

Let’s solve the second-order differential equation correctly:

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

Step 1: Solve the Homogeneous Equation

The homogeneous equation is:

y'' + 4y = 0

The characteristic equation is:

m^2 + 4 = 0

Solving for $m$ , we get:

m = \pm 2i

Thus, the solution to the homogeneous equation is:

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

Step 2: Find the Particular Solution

Since the non-homogeneous term is $\cos(2x)$ , and the homogeneous solution already contains terms involving $\cos(2x)$ and $\sin(2x)$ , we need to modify the form of the particular solution by multiplying by $x$ .

y_p = x \left( A \cos(2x) + B \sin(2x) \right)

Step 3: First Derivative of $y_p(x)$

Given:

y_p(x) = x(A \cos(2x) + B \sin(2x))

Now, differentiate using the product rule:

y_p' = (1)(A \cos(2x) + B \sin(2x)) + x(-2A \sin(2x) + 2B \cos(2x))

Simplifying:

y_p' = A \cos(2x) + B \sin(2x) - 2A x \sin(2x) + 2B x \cos(2x)

Step 2: Second Derivative of $y_p(x)$

Now differentiate $y_p'(x)$ again using the product rule:

y_p'' = \frac{d}{dx} \left(A \cos(2x) + B \sin(2x) - 2A x \sin(2x) + 2B x \cos(2x)\right)

Differentiate each term:

$\frac{d}{dx}{A\cos{(2x)}} = -2A\sin{(2x)}$
$\frac{d}{dx}(B \sin(2x)) = 2B \cos(2x)$
$\frac{d}{dx}(-2A x \sin(2x)) = -2A \sin(2x) - 4A x \cos(2x)$ (product rule)
$\frac{d}{dx}(2B x \cos(2x)) = 2B \cos(2x) - 4B x \sin(2x)$ (product rule)

Now, combine these results:

y_p'' = -2A \sin(2x) + 2B \cos(2x) - 2A \sin(2x) - 4A x \cos(2x) + 2B \cos(2x) - 4B x \sin(2x)

Simplify:

y_p'' = -4A \sin(2x) - 4A x \cos(2x) + 4B \cos(2x) - 4B x \sin(2x)

Step 3: Substitute into the Original Equation

Now substitute $y_p(x)$ , $y_p'(x)$ , and $y_p''(x)$ into the original equation $y'' + 4y = \cos(2x)$ :

(-4A \sin(2x) - 4A x \cos(2x) + 4B \cos(2x) - 4B x \sin(2x)) + 4(A x \cos(2x) + B x \sin(2x)) = \cos(2x)

The terms involving $x$ cancel out:

-4A \sin(2x) + 4B \cos(2x) = \cos(2x)

Now, equate the coefficients of $\sin(2x)$ and $\cos(2x)$ :

-4A = 0 \quad \Rightarrow \quad A = 0

4B = 1 \quad \Rightarrow \quad B = \frac{1}{4}

Step 4: Particular Solution

Thus, the particular solution is:

y_p(x) = \frac{1}{4} x \sin(2x)

Step 5: General Solution

The general solution is the sum of the homogeneous and particular solutions:

y(x) = C_1 \cos(2x) + C_2 \sin(2x) + \frac{1}{4} x \sin(2x)

2.12. Example 12

y'' - 2y' + y = e^{x}

Solution:

1. Solve the Homogeneous Equation:

y'' - 2y' + y = 0 \quad \Rightarrow \quad (m - 1)^2 = 0 \quad \Rightarrow \quad m = 1, 1

Complementary Solution:

y_c = (C_1 + C_2 x)e^x

2. Particular Solution:

Since $f(x) = e^{x}$ and $e^x$ is part of the complementary solution, assume $y_p = Ax^2 e^x$ .

Derivatives:

y_p' = 2Axe^x + Ax^2e^x, \quad y_p'' = 2Ae^x + 4Ax e^x + Ax^2 e^x

Substitute into the equation:

(2Ae^x + 4Axe^x + Ax^2e^x) - 2(2Axe^x + Ax^2 e^x) + A x^2 e^x = e^x

Simplify:

2Ae^x = e^x \quad \Rightarrow \quad A = \frac{1}{2}

y_p = \frac{1}{2}x^2 e^x

3. General Solution:

y = y_c + y_p = (C_1 + C_2 x)e^x + \frac{1}{2}x^2 e^x

3. Practice Questions

3.1. Practice Question Set 1

Find the general solution of the following equations

$\frac{d^2y}{dx^2}+2\frac{dy}{dx}-8y=2$
$\frac{d^2y}{dx^2}+3\frac{dy}{dx}-4y = 2e^{3x}$
$\frac{d^2y}{dx^2}-\frac{dy}{dx}-6y=3\cos{x}$
$\frac{d^2y}{dx^2}-2\frac{dy}{dx}+2y=2x^2-1$

Not sure about your answers? Don't worry; click the link below to check the video where the answers are explained.

Practice Questions Set 1 Discussion

3.2. Practice Question Set 2

Find the general solution of the following equations

$\frac{d^2y}{dx^2} - 4\frac{dy}{dx} +3y = 3e^{3x}$
$\frac{d^2y}{dx^2} + 5\frac{dy}{dx} + 6y = 10e^{-2x}$
$\frac{d^2y}{dx^2} + 4y = \sin(2x)$
$\frac{d^2y}{dx^2} + 3\frac{dy}{dx} + 2y = 5$

Not sure about your answers? Don't worry, click the link below to check the video where answers are explained.

Practice Questions Set 2 Discussion

Remember, the key to success in ODEs is practice and persistence. The examples and practice problems provided here are just the beginning—keep solving more problems to strengthen your understanding.

Stay tuned as we continue to build on our understanding of differential equations.

Related Articles

15. Linear non-homogeneous second-order ODEs.(Constant Coefficients)

1. Method of Undetermined Coefficients

1.1. Forcing Functions and Undetermined Coefficients

1.2. Characteristic Equation and Roots

1.3. Particular Solution and Forcing Functions

1.4. Problem Solving: Method of Undetermined Coefficients

2. Examples:

2.1. Example 1

2.2. Example 2

2.3. Example 3: Non-Homogeneous Equation with Multiplicity 1

2.4. Example 4: Non-Homogeneous Equation with Multiplicity 2

2.5. Example 5

2.6. Example 6

2.7. Example 7

2.8. Example 8

2.9. Example 9

2.10. Example 10

2.11. Example 11

2.12. Example 12

3. Practice Questions

3.1. Practice Question Set 1

3.2. Practice Question Set 2