Answer:
dx/dt=5x−3
Step-by-step explanation:
Example 1
Solve the ordinary differential equation (ODE)
dxdt=5x−3
for x(t).
Solution: Using the shortcut method outlined in the introduction to ODEs, we multiply through by dt and divide through by 5x−3:
dx5x−3=dt.
We integrate both sides
∫dx5x−315log|5x−3|5x−3x=∫dt=t+C1=±exp(5t+5C1)=±15exp(5t+5C1)+3/5.
Letting C=15exp(5C1), we can write the solution as
x(t)=Ce5t+35.
We check to see that x(t) satisfies the ODE:
dxdt=5Ce5t5x−3=5Ce5t+3−3=5Ce5t.
Both expressions are equal, verifying our solution.
Example 2
Solve the ODE combined with initial condition:
dxdtx(2)=5x−3=1.
Solution: This is the same ODE as example 1, with solution
x(t)=Ce5t+35.
We just need to use the initial condition x(2)=1 to determine C.
C must satisfy
1=Ce5⋅2+35,
so it must be
C=25e−10.
Our solution is
x(t)=25e5(t−2)+35.
You can verify that x(2)=1.
Example 3
Solve the ODE with initial condition:
dydxy(2)=7y2x3=3.
Solution: We multiply both sides of the ODE by dx, divide both sides by y2, and integrate:
∫y−2dy−y−1y=∫7x3dx=74x4+C=−174x4+C.
The general solution is
y(x)=−174x4+C.
Verify the solution:
dydx=ddx(−174x4+C)=7x3(74x4+C)2.
Given our solution for y, we know that
y(x)2=(−174x4+C)2=1(74x4+C)2.
Therefore, we see that indeed
dydx=7x3(74x4+C)2=7x3y2.
The solution satisfies the ODE.
To determine the constant C, we plug the solution into the equation for the initial conditions y(2)=3:
3=−17424+C.
The constant C is
C=−2813=−853,
and the final solution is
y(x)=−174x4−853.
