If ` u=A e ^(mx) +Be^(nx ) ,then y_2-( m+n) y_1+ mn y=`

Options

A. m
B. n
C. 1
D. 0

Answer

To solve the problem, we start with the given function:

\[ u = A e^{mx} + B e^{nx} \]

We need to find \( y_2 - (m+n) y_1 + mn y \), where \( y = u \), \( y_1 = \frac{dy}{dx} \), and \( y_2 = \frac{d^2y}{dx^2} \).

### Step 1: Calculate \( y_1 \)

First, we differentiate \( u \) with respect to \( x \):

\[ y_1 = \frac{du}{dx} = \frac{d}{dx}(A e^{mx} + B e^{nx}) \]

Using the chain rule, we get:

\[ y_1 = A m e^{mx} + B n e^{nx} \]

### Step 2: Calculate \( y_2 \)

Next, we differentiate \( y_1 \) to find \( y_2 \):

\[ y_2 = \frac{d^2u}{dx^2} = \frac{d}{dx}(A m e^{mx} + B n e^{nx}) \]

Again using the chain rule, we get:

\[ y_2 = A m^2 e^{mx} + B n^2 e^{nx} \]

### Step 3: Substitute \( y_1 \), \( y_2 \), and \( y \) into the expression

Now we substitute \( y_1 \), \( y_2 \), and \( y \) into the expression \( y_2 - (m+n) y_1 + mn y \):

\[ y_2 - (m+n) y_1 + mn y = (A m^2 e^{mx} + B n^2 e^{nx}) - (m+n)(A m e^{mx} + B n e^{nx}) + mn(A e^{mx} + B e^{nx}) \]

### Step 4: Expand the expression

Expanding the expression gives:

\[ = A m^2 e^{mx} + B n^2 e^{nx} - (m+n)(A m e^{mx}) - (m+n)(B n e^{nx}) + mn(A e^{mx}) + mn(B e^{nx}) \]

### Step 5: Group like terms

Grouping the terms for \( e^{mx} \) and \( e^{nx} \):

For \( e^{mx} \):

\[ A m^2 e^{mx} - (m+n) A m e^{mx} + mn A e^{mx} \]

Factoring out \( A e^{mx} \):

\[ A e^{mx} (m^2 - (m+n)m + mn) \]

For \( e^{nx} \):

\[ B n^2 e^{nx} - (m+n) B n e^{nx} + mn B e^{nx} \]

Factoring out \( B e^{nx} \):

\[ B e^{nx} (n^2 - (m+n)n + mn) \]

### Step 6: Simplify the expressions

Now we simplify both expressions:

1. For \( A e^{mx} \):

\[ m^2 - (m^2 + mn) + mn = 0 \]

2. For \( B e^{nx} \):

\[ n^2 - (mn + n^2) + mn = 0 \]

Thus, both expressions simplify to zero:

\[ A e^{mx} (0) + B e^{nx} (0) = 0 \]

### Final Result

Therefore, we conclude:

\[ y_2 - (m+n) y_1 + mn y = 0 \]