Note: Use partial fractions when solving
Let us assume the Laplace transform of y(t), L(y(t))=Y(s) then, we have L(y'(t))= sY(s)-y(0) and L(y"(t))= s2Y(s)-sy(0)-y'(0) . By putting the initial values y(0)=-1 and y'(0)=2 and then taking the laplace transform both sides of the given differential equation, we will get the expression of Y(s).
Then, using the Method of Partial Fraction, we will get expression of Y(s) in partial fraction form. By taking the inverse Laplace transform on both sides of Y(s), and using the rule L-1( s/(s2 + b2))= Cos (bt) and L-1(1/(s+a))= e-at
The step by step explanatory solution is provided below.
Note: Use partial fractions when solving Use the Laplace transform to solve the following initial-value problem....
17. Use the Laplace transform to solve the initial value problem: y" + 4y' + 4y = 2e-, y(0) = 1, (O) = 3. 18. Use the Laplace transform to solve the initial value problem: 4y" – 4y + 5y = 4 sin(t) – 4 cos(1), y(0) = 0, y(0) = 11/17.
Use the Laplace transform to solve the initial value problem: y' + 4y = cos(2t), y(0) = 0, y'(0 = 1.
(1 point) Use the Laplace transform to solve the following initial value problem x, = 10x + 4y, y=-6x + e4, x(0) = 0, y(0) = 0 Let x(s) L {x(t)) , and Y(s) = L {y(t)) Find the expressions you obtain by taking the Laplace transform of both differential equations and solving for Y(s) and X(s): S)E Y(s) = Find the partial fraction decomposition of X(s) and Y(s) and their inverse Laplace transforms to find the solution of the...
Please show work Question 14 5 pts Use the Laplace transform to solve the given initial-value problem. y" + 4y=f(t – 2), y(0) = 1, y (0) = 0 Oy(t) = cos(2t) + U (t – 2) · sin[2(t – 2)] Oy(t) = {U (t – 2) sin(2t) Oy(t) = {U (t – 2) sin(2(t – 2)] Oy(t) = cos(2t) + U (t – 2) sin(2t)
use the Laplace transform to solve the given initial value problem: Only problem 4,8 and 12 please 4. y" – 4y' + 4y = 0; y(0) = 1, y'(0) = 1 5. y" – 2y' + 4y = 0; y(0) = 2, y'(0) = 0 Σ Answer Solution = e 6. y" + 4y' + 297 - 2t sin 5t; y(0) = 5, 7. y" +12y = cos 2t, 22 # 4; y(0) = 1, : > Answer > Solution...
(4 points) Use the Laplace transform to solve the following initial value problem: y" – 2y + 5y = 0 y(0) = 0, y'(0) = 8 First, using Y for the Laplace transform of y(t), i.e., Y = L{y(t)}| find the equation you get by taking the Laplace transform of the differential equation = 01 Now solve for Y(3) By completing the square in the denominator and inverting the transform, find g(t) =
In this exercise we will use the Laplace transform to solve the following initial value problem: y"-2y'+ 17y-17, y(0)=0, y'(0)=1 (1) First, using Y for the Laplace transform of y(t), i.e., Y =L(y(t)), find the equation obtained by taking the Laplace transform of the initial value problem (2) Next solve for Y= (3) Finally apply the inverse Laplace transform to find y(t)
Use the Laplace transform to solve the given initial-value problem. y" + 6y' + 5y = 0, y(0) = 1, y'(O) = 0 y(t) =
Use the Laplace transform to solve the following initial value problem: y" + 4y = f(t), y(0) = 0, y'(0) = 1 where f(t) = { if 0 <t <a sint if a <t< oo
1 point) Use the Laplace transform to solve the following initial value problem: y" - 9y' + 18y-0, y(0) -3, y' (0) 3 (1) First, using Y for the Laplace transform of y(t), i.e., Y-C00), find the equation you get by taking the Laplace transform of the differential equation to obtain (2) Next solve for Y (3) Now write the above answer in its partial fraction form, Y- (NOTE: the order that you enter your answers matter so you must...