solve the following differential equations
dx/dt - 7x + y = 0 -->(1)
dy/dt - 2x - 5y = 0 -->(2)
(2)/(1)
dy/dx = (2x+5y)/(7x-y)
y=vx
Etc.
Is this a correct way to solve this one?
so far, so good.
Yes, substituting y = vx is a correct method for solving this system of differential equations. By substituting this expression into the equations, we can simplify and solve for the values of x and v.
Let's proceed step-by-step:
1. Start with the given equations:
dx/dt - 7x + y = 0 ...(1)
dy/dt - 2x - 5y = 0 ...(2)
2. Substitute y = vx into both equations:
dx/dt - 7x + vx = 0 ...(1')
v(dx/dt) - 2x - 5vx = 0 ...(2')
3. Rearrange Equation (2') to isolate dx/dt:
dx/dt = (2x + 5vx)/(v + 7) ...(3)
4. Substitute Equation (3) back into Equation (1'). This will eliminate dx/dt:
(2x + 5vx)/(v + 7) - 7x + vx = 0
5. Simplify the equation:
2x + 5vx - 7x(v + 7) + vx(v + 7) = 0
2x + 5vx - 7vx - 49x + 7xv + 49v + vx^2 + 7vx = 0
2x - 49x + 49v + x^2v + 7vx^2 = 0
6. Combine like terms:
x^2v + 7vx^2 - 47x + 49v = 0
7. Factor out x:
x(xv + 7v - 47) + 49v = 0
8. Set each term equal to zero:
x = 0 ...(4)
xv + 7v - 47 = 0 ...(5)
49v = 0 ...(6)
9. Solve Equation (6) for v. We have v = 0. Substituting back into Equation (5):
(0) + 7(0) - 47 = 0
-47 = 0 (This is not possible)
10. Since Equation (6) gives us v = 0, return to Equation (4) to solve for x.
x = 0
11. Therefore, the solution to the system of differential equations is x = 0 and y = 0.
So, substituting y = vx did lead us to the solution x = 0 and y = 0.
Yes, your approach is correct. To solve the given system of differential equations, you can follow these steps:
Step 1: Solve equation (1) for y in terms of x by isolating y:
dx/dt - 7x + y = 0
y = 7x - dx/dt ---> (3)
Step 2: Differentiate equation (3) with respect to x to find dy/dx:
dy/dx = d/dx (7x - dx/dt)
dy/dx = 7 - d²x/dt² ---> (4)
Step 3: Substitute equation (4) into equation (2):
dy/dt - 2x - 5y = 0
7 - d²x/dt² - 2x - 5(7x - dx/dt) = 0 (Substituting dy/dx from (4))
7 - d²x/dt² - 2x - 35x + 5(dx/dt) = 0
-d²x/dt² - 37x + 5(dx/dt) = -7 ---> (5)
Step 4: Differentiate equation (1) with respect to t to find dx/dt:
d²x/dt² - 7(dx/dt) + dy/dt = 0
d²x/dt² - 7(dx/dt) + 2x + 5y = 0 (Substituting dy/dt from (2))
d²x/dt² - 7(dx/dt) + 2x + 5(7x - dx/dt) = 0
d²x/dt² - 7(dx/dt) + 2x + 35x - 5(dx/dt) = 0
d²x/dt² - 12(dx/dt) + 37x = 0
Step 5: Substitute equation (5) into equation (1):
-d²x/dt² - 37x + 5(dx/dt) = -7 (From equation (5))
d²x/dt² + 37x - 5(dx/dt) = 7 (Change signs)
d²x/dt² + 37x = 5(dx/dt) + 7 ---> (6)
Step 6: Combine equations (5) and (6) to solve for dx/dt:
From equation (5): -d²x/dt² - 37x + 5(dx/dt) = -7 ---> (5)
5(dx/dt) = -7 + d²x/dt² + 37x (Move terms around)
dx/dt = (-7 + d²x/dt² + 37x)/5 (Divide by 5)
Step 7: Substitute the value of dx/dt from (6) into (5):
d²x/dt² + 37x = 5(dx/dt) + 7 ---> (6)
d²x/dt² + 37x = 5((-7 + d²x/dt² + 37x)/5) + 7 (Substituting dx/dt from Step 6)
d²x/dt² + 37x = -7 + d²x/dt² + 37x + 7
Simplifying, we get: 0 = 0
Since the equation 0 = 0 is an identity, it means that dx/dt and x can have any values, and y can be expressed as y = vx, where v is an arbitrary constant. Therefore, the general solution to the given system of differential equations is x(t) = x, y(t) = vx, where v is any constant.