Answer :
Sure, let's solve the given differential equation step-by-step to determine the largest value of [tex]\( k \)[/tex].
The differential equation given is:
[tex]\[ \frac{d^2 y}{dx^2} - 7 \frac{dy}{dx} + 6y = 0. \][/tex]
We will use the method of assuming a solution in the form:
[tex]\[ y = e^{kx}. \][/tex]
Then,
[tex]\[ \frac{dy}{dx} = ke^{kx} \][/tex]
and
[tex]\[ \frac{d^2 y}{dx^2} = k^2 e^{kx}. \][/tex]
Substitute [tex]\( y \)[/tex], [tex]\( \frac{dy}{dx} \)[/tex], and [tex]\( \frac{d^2 y}{dx^2} \)[/tex] into the differential equation:
[tex]\[ k^2 e^{kx} - 7k e^{kx} + 6 e^{kx} = 0. \][/tex]
Factor out [tex]\( e^{kx} \)[/tex] (noting that [tex]\( e^{kx} \neq 0 \)[/tex]):
[tex]\[ e^{kx} (k^2 - 7k + 6) = 0. \][/tex]
This gives us the quadratic equation:
[tex]\[ k^2 - 7k + 6 = 0. \][/tex]
Now, we need to solve this quadratic equation. The general form of a quadratic equation is [tex]\( ax^2 + bx + c = 0 \)[/tex]. For our equation, [tex]\( a = 1 \)[/tex], [tex]\( b = -7 \)[/tex], and [tex]\( c = 6 \)[/tex].
We calculate the discriminant ([tex]\( \Delta \)[/tex]) of the quadratic equation:
[tex]\[ \Delta = b^2 - 4ac. \][/tex]
Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ \Delta = (-7)^2 - 4 \cdot 1 \cdot 6 = 49 - 24 = 25. \][/tex]
Next, we find the square root of the discriminant:
[tex]\[ \sqrt{\Delta} = \sqrt{25} = 5. \][/tex]
Using the quadratic formula [tex]\( k = \frac{-b \pm \sqrt{\Delta}}{2a} \)[/tex]:
[tex]\[ k = \frac{-(-7) \pm 5}{2 \cdot 1} = \frac{7 \pm 5}{2}. \][/tex]
This gives us two solutions for [tex]\( k \)[/tex]:
[tex]\[ k_1 = \frac{7 + 5}{2} = \frac{12}{2} = 6 \][/tex]
and
[tex]\[ k_2 = \frac{7 - 5}{2} = \frac{2}{2} = 1. \][/tex]
Therefore, the values of [tex]\( k \)[/tex] are [tex]\( k_1 = 6 \)[/tex] and [tex]\( k_2 = 1 \)[/tex].
The largest value of [tex]\( k \)[/tex] is:
[tex]\[ \boxed{6} \][/tex]
The differential equation given is:
[tex]\[ \frac{d^2 y}{dx^2} - 7 \frac{dy}{dx} + 6y = 0. \][/tex]
We will use the method of assuming a solution in the form:
[tex]\[ y = e^{kx}. \][/tex]
Then,
[tex]\[ \frac{dy}{dx} = ke^{kx} \][/tex]
and
[tex]\[ \frac{d^2 y}{dx^2} = k^2 e^{kx}. \][/tex]
Substitute [tex]\( y \)[/tex], [tex]\( \frac{dy}{dx} \)[/tex], and [tex]\( \frac{d^2 y}{dx^2} \)[/tex] into the differential equation:
[tex]\[ k^2 e^{kx} - 7k e^{kx} + 6 e^{kx} = 0. \][/tex]
Factor out [tex]\( e^{kx} \)[/tex] (noting that [tex]\( e^{kx} \neq 0 \)[/tex]):
[tex]\[ e^{kx} (k^2 - 7k + 6) = 0. \][/tex]
This gives us the quadratic equation:
[tex]\[ k^2 - 7k + 6 = 0. \][/tex]
Now, we need to solve this quadratic equation. The general form of a quadratic equation is [tex]\( ax^2 + bx + c = 0 \)[/tex]. For our equation, [tex]\( a = 1 \)[/tex], [tex]\( b = -7 \)[/tex], and [tex]\( c = 6 \)[/tex].
We calculate the discriminant ([tex]\( \Delta \)[/tex]) of the quadratic equation:
[tex]\[ \Delta = b^2 - 4ac. \][/tex]
Substituting the values of [tex]\( a \)[/tex], [tex]\( b \)[/tex], and [tex]\( c \)[/tex]:
[tex]\[ \Delta = (-7)^2 - 4 \cdot 1 \cdot 6 = 49 - 24 = 25. \][/tex]
Next, we find the square root of the discriminant:
[tex]\[ \sqrt{\Delta} = \sqrt{25} = 5. \][/tex]
Using the quadratic formula [tex]\( k = \frac{-b \pm \sqrt{\Delta}}{2a} \)[/tex]:
[tex]\[ k = \frac{-(-7) \pm 5}{2 \cdot 1} = \frac{7 \pm 5}{2}. \][/tex]
This gives us two solutions for [tex]\( k \)[/tex]:
[tex]\[ k_1 = \frac{7 + 5}{2} = \frac{12}{2} = 6 \][/tex]
and
[tex]\[ k_2 = \frac{7 - 5}{2} = \frac{2}{2} = 1. \][/tex]
Therefore, the values of [tex]\( k \)[/tex] are [tex]\( k_1 = 6 \)[/tex] and [tex]\( k_2 = 1 \)[/tex].
The largest value of [tex]\( k \)[/tex] is:
[tex]\[ \boxed{6} \][/tex]
