Two solutions to y'' – 2y' – 35y = 0 are yı = e, y2 = e -5t a) find the wronskian. w = 0 preview b) find the solution satisfying the initial conditions y(0) = – 7, y'(0) = 23 y = ( preview

a.

b.

Step-by-step explanation:

We have a differential equation

y''-2 y'-35 y=0

Auxillary equation

By factorization method we are  finding the solution

Substitute each factor equal to zero

D-7=0  and D+5=0

D=7  and D=-5

Therefore ,

General solution is

Let

We have to find Wronskian

Substitute values then we get

a.

We are given that y(0)=-7 and y'(0)=23

Substitute the value in general solution the we get

....(equation I)

......(equation II)

Equation I is multiply by 5 then we subtract equation II from equation I

Using elimination method we eliminate

Then we get

Substitute the value of in  I equation then we get

Hence, the general solution is

b.

step-by-step explanation:

here's what happens:

1) the graph of f   = x^2 is shifted 2 units to the right.

2) the resulting graph is reflected in the x-axis, so that the parabola now opens down.

3) this final result is transformed up by 5 units.

if the pizza is divided equally into 2 equal halves then you would get

πr^2/2

where r is the radius of the pizza.

happy to

pls mark as brainliest