The differential equation of the family of curves `y=e^(2x)(a cos x+b sin x)` where, a and b are arbitrary constants, is given by

Options

A. `y_(2)-4y_(1)+5y=0`
B. `2y_(2)-y_(1)+5y=0`
C. `y_(2)+4y_(1)-5y=0`
D. `y_(2)-2y_(1)+5y=0`

Answer

To find the differential equation of the family of curves given by \( y = e^{2x}(a \cos x + b \sin x) \), where \( a \) and \( b \) are arbitrary constants, we will follow these steps:

### Step 1: Differentiate \( y \) with respect to \( x \)

We start with the function:
\[ y = e^{2x}(a \cos x + b \sin x) \]

Using the product rule for differentiation, which states that \( (uv)' = u'v + uv' \), we differentiate \( y \):

Let \( u = e^{2x} \) and \( v = a \cos x + b \sin x \).

1. Differentiate \( u \):
\[ u' = \frac{d}{dx}(e^{2x}) = 2e^{2x} \]

2. Differentiate \( v \):
\[ v' = \frac{d}{dx}(a \cos x + b \sin x) = -a \sin x + b \cos x \]

Now, applying the product rule:
\[ \frac{dy}{dx} = u'v + uv' = 2e^{2x}(a \cos x + b \sin x) + e^{2x}(-a \sin x + b \cos x) \]

Factor out \( e^{2x} \):
\[ \frac{dy}{dx} = e^{2x} \left( 2(a \cos x + b \sin x) + (-a \sin x + b \cos x) \right) \]
\[ = e^{2x} \left( (2a + b) \cos x + (2b - a) \sin x \right) \]

### Step 2: Differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \)

Now, we differentiate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = e^{2x} \left( (2a + b) \cos x + (2b - a) \sin x \right) \]

Using the product rule again:
Let \( u = e^{2x} \) and \( v = (2a + b) \cos x + (2b - a) \sin x \).

1. Differentiate \( u \):
\[ u' = 2e^{2x} \]

2. Differentiate \( v \):
\[ v' = \frac{d}{dx}((2a + b) \cos x + (2b - a) \sin x) = -(2a + b) \sin x + (2b - a) \cos x \]

Now applying the product rule:
\[ \frac{d^2y}{dx^2} = u'v + uv' = 2e^{2x} \left( (2a + b) \cos x + (2b - a) \sin x \right) + e^{2x} \left( -(2a + b) \sin x + (2b - a) \cos x \right) \]

Factor out \( e^{2x} \):
\[ \frac{d^2y}{dx^2} = e^{2x} \left( 2((2a + b) \cos x + (2b - a) \sin x) + (-(2a + b) \sin x + (2b - a) \cos x) \right) \]

### Step 3: Substitute \( y \) and \( \frac{dy}{dx} \) into the equation

Now we can express \( \frac{d^2y}{dx^2} \) in terms of \( y \) and \( \frac{dy}{dx} \).

From the previous steps, we can derive that:
\[ \frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 5y = 0 \]

### Final Differential Equation

Thus, the required differential equation of the family of curves is:
\[ \frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 5y = 0 \]

---