Question

Consider the differential equation,

L[y] = y'' + p(t)y' + q(t)y = 0,

(1) whose coefficients p and q are continuous on some open interval I. Choose some point t0 in I. Let y1 be the solution of equation (1) that also satisfies the initial conditions y(t0) = 1, y'(t0) = 0, and let y2 be the solution of equation (1) that satisfies the initial conditions y(t0) = 0, y'(t0) = 1. Then y1 and y2 form a fundamental set of solutions of equation (1). Find the fundamental set of solutions specified by the theorem above for the given differential equation and initial point. y'' + 7y' − 8y = 0, t0 = 0

Answer 1

Answer:

The two solutions are given as $y_1(t)=\dfrac{1}{9}e^{-8t}+\dfrac{8}{9}e^{t}$ and $y_2(t)=\dfrac{-1}{9}e^{-8t}+\dfrac{1}{9}e^{t}$

Step-by-step explanation:

As the given equation is

$y''+7y'-8y=0$

So the corresponding equation is given as

$m^2+7m-8=0$

Solving this equation yields the value of m as

$(m+8)(m-1)=0\\m=-8, m=1$

Now the equation is given as

$y(t)=C_1e^{m_1t}+C_2e^{m_2t}$

Here m1=-8, m2=1 so

$y(t)=C_1e^{-8t}+C_2e^{t}$

The derivative is given as

$y'(t)=-8C_1e^{-8t}+C_2e^{t}$

Now for the first case y(t=0)=1, y'(t=0)=0

$y(t=0)=C_1e^{-8*0}+C_2e^{0}\\1=C_1+C_2\\\\y'(t=0)=-8C_1e^{-8*0}+C_2e^{0}\\0=-8C_1+C_2$

So the two equation of co-efficient are given as

$C_1+C_2=1\\-8C_1+C_2=0$

Solving the equation yield

$C_1=1/9 \\C_2=8/9$

So the function is given as

$y_1(t)=\dfrac{1}{9}e^{-8t}+\dfrac{8}{9}e^{t}$

Now for the second case y(t=0)=0, y'(t=0)=1

$y(t=0)=C_1e^{-8*0}+C_2e^{0}\\0=C_1+C_2\\\\y'(t=0)=-8C_1e^{-8*0}+C_2e^{0}\\1=-8C_1+C_2$

So the two equation of co-efficient are given as

$C_1+C_2=0\\-8C_1+C_2=1$

Solving the equation yield

$C_1=-1/9 \\C_2=1/9$

So the function is given as

$y_2(t)=\dfrac{-1}{9}e^{-8t}+\dfrac{1}{9}e^{t}$

So the two solutions are given as $y_1(t)=\dfrac{1}{9}e^{-8t}+\dfrac{8}{9}e^{t}$ and $y_2(t)=\dfrac{-1}{9}e^{-8t}+\dfrac{1}{9}e^{t}$