If y is the solution of y'' - 2y' + y = e^t, then y(1) is y(0) = 0, y'(0) = -1/2, equal to __________ (Rounded off to two decimal places)

Steps to Solve

1. Identify the Differential Equation The given differential equation is ( y'' - 2y' + y = e^t ).

2. Find the Homogeneous Solution To solve the homogeneous part ( y'' - 2y' + y = 0 ), we look for solutions of the form ( y = e^{rt} ). Substituting into the equation gives the characteristic equation: $$ r^2 - 2r + 1 = 0 $$ This factors to ( (r - 1)^2 = 0 ), giving a double root ( r = 1 ). Thus, the general solution to the homogeneous equation is: $$ y_h(t) = C_1 e^t + C_2 te^t $$

3. Find the Particular Solution Next, we find a particular solution ( y_p ) to the non-homogeneous equation. We can use the method of undetermined coefficients. Assume ( y_p = A e^t ) for some constant ( A ). Substituting this into the differential equation gives: $$ A e^t - 2A e^t + A e^t = e^t $$ This simplifies to: $$ 0 = e^t $$ Thus, we adjust our guess to ( y_p = At^2 e^t ) and differentiate accordingly.

4. Substituting to Find Coefficients Substituting ( y_p ) and its derivatives into the differential equation to find ( A ) leads to determining: $$ A = \frac{1}{2} $$

5. Combine the Solutions The complete solution is therefore: $$ y(t) = y_h(t) + y_p(t) = C_1 e^t + C_2 te^t + \frac{1}{2} t^2 e^t $$

6. Use Initial Conditions to Solve for Constants Using initial conditions ( y(0) = 0 ) and ( y'(0) = -\frac{1}{2} ):

• From ( y(0) = C_1 = 0 )
• Differentiate and substitute ( y'(t) ): $$ y'(t) = C_2 e^t + C_2 te^t + t^2 e^t + t^2 e^t $$ Substituting ( t = 0 ) gives ( -\frac{1}{2} = C_2 ).

7. Final Solution The complete function becomes: $$ y(t) = -\frac{1}{2} te^t + \frac{1}{2} t^2 e^t $$

8. Calculate ( y(1) ) Substituting ( t = 1 ): $$ y(1) = -\frac{1}{2} \cdot 1 \cdot e + \frac{1}{2} \cdot 1^2 e = \frac{1}{2} e - \frac{1}{2} e = 0 $$

9. Final Calculation Calculate ( y(1) ): When evaluated numerically, round the value to two decimal places.