Final answer:
To solve the system of first-order ODEs x′ − 2y′ + x = 10 cos t and x′ − y′ + 6y = 0, we can use the method of elimination. By multiplying the first equation by 3 and subtracting the second equation, we can eliminate x and y'. Simplifying the resulting equation and assuming a solution of the form x = A cos t + B sin t, we can solve for A and B to find the values of x and y.
Step-by-step explanation:
To solve the system of first-order ODEs x′ − 2y′ + x = 10 cos t and x′ − y′ + 6y = 0, we can use the method of elimination.
First, we can multiply the first equation by 3 to make the coefficients of y' the same in both equations. This gives us 3x′ − 6y′ + 3x = 30 cos t.
Now, we can subtract the second equation from this new equation to eliminate x and y'. This gives us:
2y′ − 5x = 30 cos t.
From here, we can solve for y′ by dividing both sides of the equation by 2: y′ = (30/2) cos t - (5/2) x.
Now that we have the value of y′, we can substitute it back into one of the original equations to solve for x′. Let's use the first equation: x′ − 2((30/2) cos t - (5/2) x) + x = 10 cos t.
Simplifying this equation gives us x′ − 30 cos t + 5x = 0.
Now we have a second-order ODE in terms of x. To solve this equation, we can assume a solution of the form x = A cos t + B sin t, where A and B are constants.
Substituting this solution into the equation gives us:
(-A sin t - B cos t) − 30 cos t + 5(A cos t + B sin t) = 0.
Now, we can group like terms and solve for A and B:
-A + 5A + 30 cos t - B − 5B cos t = 0.
Combining like terms gives us: (4A - 5B) cos t + (-A - 5B) sin t + 30 cos t = 0.
Since the coefficients of cos t and sin t must be equal to zero for all t, we can set them equal.
From this, we get the following system of equations:
4A - 5B + 30 = 0 and -A - 5B = 0.
Solving this system of equations gives us A = 6 and B = -1.2.
Therefore, the solution to the system of first-order ODEs is:
x = 6 cos t - 1.2 sin t and y = (30/2) cos t - (5/2)(6 cos t - 1.2 sin t).