

y'' + 2y' + y = x + e^(-x) , y(0)=0 , y'(0)=0 Solve the following 2nd ODE and find y(x)
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U. + 2y + y + 1 -e: y(0) = 0, y'(o) - 2 In Problems 31-36, determine the form of a particular solution for the differential equation. Do not solve. 31. y" + y = sin : + i cos + + 10' 32. y" - y = 2+ + te? + 1221 x" - x' - 2x = e' cos - + cost y" + 5y' + 6y = sin t - cos 2t 35. y" –...
Find a solution
10. y" – 2y' + 2y = 2x, y(0) = 4, y'0) = 8.
Show all work for each problem. 1. (15 pts) y"-2y'+2y = 2x, y(0) = 4, y"0) = 8, y, =ce" cosx+c,e' sin x, y, = x+1. Find a solution satisfying the given initial conditions.
2y + y + 2y = g(t), (O) = 0, y'(0) = 0 where g) 5 St<20 10, 0<t<5 and t > 20
4. Find the solution to the differential equation y"+2y'+ 2y-S(t-) y(0) 0, y (0)-0 and graph it.
Solve the Following: 2y'' + y'+ 2y = u5(t) − u20(t) y(0) = −1 y 0 (0) = 3
1. The function: y, = e' is a solution of the homogeneous linear equation: y"-2y'+ y = 0. Use Reduction of Order to find a second linearly independent solution, then write the general solution for the differential equation: y" - 2y'+y=0
1) y'' -2y'+y=xE^x,
y(0)=y'(0)=0 Solve the initial value problem using the Laplace
transform.
y" – 2y + y = xe*, y(0) = y'(0) =
2.5. y"yuT/2 (t) 8(t-) - u3/2(t); y(0) = 0, y'(0) = 0 2.6. y" 2y 2y cos t6(t T/2); y(0) = 0, y'(0) = 0