## ANS:

## Finding the Particular Integrals:

**a) y” + y = cos(2x – 1)**

**Homogeneous Solution:**

First, solve the homogeneous equation: y” + y = 0. This is a second-order linear homogeneous equation with constant coefficients. The characteristic equation is r^2 + 1 = 0, which has roots r = ±i. Therefore, the homogeneous solution is:

```
y_h(x) = C₁ cos(x) + C₂ sin(x)
```

**Particular Solution:**

Since the non-homogeneous term is cos(2x – 1), we can use the method of undetermined coefficients. We assume a particular solution of the form:

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

Differentiate y_p(x) twice to find y_p”(x):

```
y_p''(x) = -4A cos(2x - 1) + 4B sin(2x - 1)
```

Substitute y_p(x) and y_p”(x) into the original equation:

```
(-4A cos(2x - 1) + 4B sin(2x - 1)) + (A cos(2x - 1) + B sin(2x - 1)) = cos(2x - 1)
```

Simplify and equate coefficients:

-4A + A = 1 4B + B = 0

Solve for A and B:

A = 1/3 B = 0

Therefore, the particular solution is:

```
y_p(x) = 1/3 cos(2x - 1)
```

**General Solution:**

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

```
y(x) = C₁ cos(x) + C₂ sin(x) + 1/3 cos(2x - 1)
```

**b) y” + y’ = sin(2x)**

**Homogeneous Solution:**

The homogeneous equation is y” + y’ = 0. This is a first-order linear homogeneous equation with constant coefficients. The characteristic equation is r^2 + r = 0, which has roots r = 0, -1. Therefore, the homogeneous solution is:

```
y_h(x) = C₁ + C₂ e^(-x)
```

**Particular Solution:**

Since the non-homogeneous term is sin(2x), we can use the method of undetermined coefficients. We assume a particular solution of the form:

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

Differentiate y_p(x) and y_p'(x) to find y_p”(x):

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

Substitute y_p(x), y_p'(x), and y_p”(x) into the original equation:

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

Simplify and equate coefficients:

-4A – 2A = 0 -4B + 2B = 1

Solve for A and B:

A = 0 B = 1/2

Therefore, the particular solution is:

```
y_p(x) = 1/2 sin(2x)
```

**General Solution:**

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

`y(x) = C₁ + C₂ e^(-x) + 1/2 sin(2x)`