Solve the initialvalue problem y 4y 4y 0 y0 1 y0 2Solut

Solve the initial-value problem, y\" - 4y\' + 4y = 0, y(0) = 1, y\'(0) = 2

Solution

The problem can be rewritten as   d²y/dx² -4*dy/dx + 4 = 0 .

The Auxiliary equation shall be m² -4m +4 =0 i.e. m= 2,2

Therefore solution is   y = (c1 + c2x).e^2x

Now the constants c1 and c2 shall be evaluated using the given initial values y(0) = 1 and y\'(0) = 2 .

Put y = 1 for x=0 in the solution y = (c1 + c2x).e^2x giving 1 = (c1+c2. 0) e^2.0 = c1 .1 è c1 = 1 .

Now Y= (1 + c2.x).e^2x   . Differentiating , dy/dx =   (0+    c2.1)e^2x + (1+c2.x).2.e^2x

As dy/dx at x = 0 is 2 ,    2 = c2.1 + 1.2.1 è c2 = 0 .

Therefore , the final solution is y = e^2x   .