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

The two solutions are given as and

Step-by-step explanation:

As the given equation is

So the corresponding equation is given as

Solving this equation yields the value of m as

Now the equation is given as

Here m1=-8, m2=1 so

The derivative is given as

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

So the two equation of co-efficient are given as

Solving the equation yield

So the function is given as

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

So the two equation of co-efficient are given as

Solving the equation yield

So the function is given as

So the two solutions are given as and

the answer is b.

step-by-step explanation:

by the trigonometric ratio tangent, tangent equals opposite divided by adjacent.

therefore, tan(50) equals ac/3

multiply each side by 3 to get: ac=3tan(50)

the answer is b.

1/3 + 1/5 = 8/15.

step-by-step explanation:

hope it