first we solve the differential equation using laplace transform and then comparing it with series RLC circuit transfer function we get value of R , L and C .
BP: Given the following differential equation, solve and design the RLC Circuit. Y + 5y +...
Solve the following differential equation with given initial conditions using the Laplace transform. y" + 5y' + 6y = ut - 1) - 5(t - 2) with y(0) -2 and y'(0) = 5. 1 AB I
Solve the given third-order differential equation by variation of parameters. y" - 5y" – y' + 5y = e** y(x) =
Question 5 < > Given the differential equation y' + 5y' + 4y = 0, y(0) = 2, y'(0) = 1 Apply the Laplace Transform and solve for Y(8) = L{y} Y(s) = Now solve the IVP by using the inverse Laplace Transform y(t) = L-'{Y(s)} g(t) =
please solve with steps and explain thanks Question 5 Given the differential equation y'' + 5y' + 4y = 0, y(0) = 2, y'(0) = 1 Apply the Laplace Transform and solve for Y(8) = L{y} Y(8) = Now solve the IVP by using the inverse Laplace Transform y(t) = L-'{Y(s)} g(t) =
Solve the differential equation given the inital conditions provided: 1' – 4y + 5y = cos&, y(0) = 0, y (0) = 1
Given: y''+5y'+6y=x solve for all representations USING MATLAB 1. Differential System 2. Impulse Response 3.La Place (transfer function) 4.block diagram 5.state equation 6. draw schematic using op-amps please show the code used in matlab
Given the differential equation y"' + 5y' – 4y = 4 sin(3t), y(0) = 2, y'(0) = -1 Apply the Laplace Transform and solve for Y(s) = L{y} Y(s) = 1
Tutorial Exercise Use the Laplace transform to solve the given initial-value problem. y' + 5y = et (0) = 2 Step 1 To use the Laplace transform to solve the given initial value problem, we first take the transform of each member of the differential equation + 6y et The strategy is that the new equation can be solved for ty) algebraically. Once solved, transforming back to an equation for gives the solution we need to the original differential equation....
differential and linear equations please please double check answers are correct Solve y" + 5y' + 6y = S(t – 2) y(0) = 0 y'(O) = 1
Given the differential equation y” + 5y' – 4y = 4 sin(3t), y(0) = 2, y'(0) = -1 Apply the Laplace Transform and solve for Y(s) = L{y} Y(s) = (293 +52 + 188 +21) (52 +58 - 4)( 92 +9)