I got an answer but not sure. I really need help on this last problem. Solve bvp : y"-3y'+2y=0 ,y(0)=1, y(3)=0

Question

I got y=(e^(2x)(e^3+2e^6)-e^x (e^6))/e^3-e^6

Answer 1

Assume a solution of the form y = e^ax

This will give a characteristic equation:

a² - 3a + 2 = 0

a = [3 ± √(9 - 4*2)]/2 = (3 ± 1)/2 = 2 or 1

Then your general solution is a linear combination of e^2x and e^x.

y = Ae^2x + Be^x

y(0) = 1 ⇒ 1 = A + B
y(3) = 0 ⇒ 0 = Ae⁶ + Be³ = e³(Ae³ + B) = 0 ⇒ Ae³ + B = 0

Solve for A and B.

A = 1 - B

(1 - B)e³ + B = 0

e³ – Be³ + B = 0

e³ + B(1 – e³) = 0

B = e³/(e³ – 1)

A = 1 – B = (e³ – 1 – e³)/(e³ – 1) = -1/(e³ – 1) = 1/(1 – e³)

y(x) = [1/(1 – e³)] e^2x + [e³/(e³ – 1)] e^x

Test the solution:

y(0) = 1/(1 – e³) + e³/(e³ – 1) = [1/(1 – e³)] (1 – e³) = 1 [good]

y(3) = [1/(1 – e³)]e⁶ + [e³/(e³ – 1)]e³ = [1/(1 – e³)](e⁶ – e⁶) = 0 [good]

Answer 2

This is a homeogenous second-order equation.

 

Let       y = e^xt

          y' = xe^xt

         y'' = x²e^xt

 

Substituting these values into the equation we get

 

e^xt(x² - 3x + 2) = 0

 

x² - 3x + 2 = 0

 

 

This is the characteristic equation.  Solve for x from this characteristic equation.

 

(x - 1)(x - 2) = 0

 

x = 1      and      x = 2

 

 

Using these x values, the general solution of the differential equation is

 

y(x) = Ae^x + Be^2x

 

 

Since we have initial value conditions, we can use them to solve for A and B.

 

y(0)= 1

 

A + B = 1

 

 

y(3) = 0

 

Ae³ + Be⁶ = 0

 

The bolded equation becomes your system of equation to solve for A and B.   So use substitution/elimination method to solve for A and B.