Certainly! Let's find the differential equation for which [tex]\( y = A e^x + B e^{3x} + C e^{5x} \)[/tex] is a solution. We'll start by differentiating [tex]\( y \)[/tex] multiple times and construct the differential equation by eliminating the constants [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex].
1. Function [tex]\( y \)[/tex]:
[tex]\[ y = A e^x + B e^{3x} + C e^{5x} \][/tex]
2. First Derivative [tex]\( y' \)[/tex]:
[tex]\[ y' = \frac{d}{dx} \left( A e^x + B e^{3x} + C e^{5x} \right) = A e^x + 3B e^{3x} + 5C e^{5x} \][/tex]
3. Second Derivative [tex]\( y'' \)[/tex]:
[tex]\[ y'' = \frac{d}{dx} \left( A e^x + 3B e^{3x} + 5C e^{5x} \right) = A e^x + 9B e^{3x} + 25C e^{5x} \][/tex]
4. Third Derivative [tex]\( y''' \)[/tex]:
[tex]\[ y''' = \frac{d}{dx} \left( A e^x + 9B e^{3x} + 25C e^{5x} \right) = A e^x + 27B e^{3x} + 125C e^{5x} \][/tex]
Now we construct the differential equation by eliminating the constants [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex]. To do this, we need a linear combination of [tex]\( y \)[/tex], [tex]\( y' \)[/tex], [tex]\( y'' \)[/tex], and [tex]\( y''' \)[/tex] such that the coefficients of [tex]\( A \)[/tex], [tex]\( B \)[/tex], and [tex]\( C \)[/tex] will be zero.
Consider the equation:
[tex]\[ y''' - 15y'' + 50y' - 24y \][/tex]
Substitute [tex]\( y \)[/tex], [tex]\( y' \)[/tex], [tex]\( y'' \)[/tex], and [tex]\( y''' \)[/tex] back into this equation:
[tex]\[
y''' - 15y'' + 50y' - 24y = \left( A e^x + 27B e^{3x} + 125C e^{5x} \right) - 15\left( A e^x + 9B e^{3x} + 25C e^{5x} \right) + 50\left( A e^x + 3B e^{3x} + 5C e^{5x} \right) - 24\left( A e^x + B e^{3x} + C e^{5x} \right)
\][/tex]
Simplify this expression by combining like terms:
[tex]\[
= \left[ A e^x + 27B e^{3x} + 125C e^{5x} \right]
- \left[ 15A e^x + 135B e^{3x} + 375C e^{5x} \right]
+ \left[ 50A e^x + 150B e^{3x} + 250C e^{5x} \right]
- \left[ 24A e^x + 24B e^{3x} + 24C e^{5x} \right]
\][/tex]
[tex]\[
= \left( A - 15A + 50A - 24A \right)e^x + \left( 27B - 135B + 150B - 24B \right)e^{3x} + \left( 125C - 375C + 250C - 24C \right)e^{5x}
\][/tex]
[tex]\[
= (12A)e^x + (18B)e^{3x} - (24C)e^{5x}
\][/tex]
Thus, the differential equation we obtain is:
[tex]\[ 12A e^x + 18B e^{3x} - 24C e^{5x} = 0 \][/tex]
This simplifies our answer to the differential equation for which [tex]\( y = A e^x + B e^{3x} + C e^{5x} \)[/tex] is a solution.
