Differential Equations

Step 1: Write the characteristic equation.
For the given ODE: Assume solution of form . Substituting:
Step 2: General solution.

Step 3: Apply initial conditions.
At , : Now derivative: At , :
Step 4: Solve for constants.
From (1): . Substitute into (2):
Step 5: Final solution.

Step 6: Evaluate at .
Oops! Let’s check again carefully. (We must recalc because I typed wrong in Final Answer earlier.) At : So final result is: Final Answer:

Step 1: Recall the definition.
A differential equation is linear if the dependent variable ( ) and its derivatives appear only in the first power and are not multiplied/divided by each other or in nonlinear functions like , , etc.
Step 2: Analyze equation P.
P: Here, the term makes the equation nonlinear (since appears inside a trigonometric function). Hence, P is nonlinear.
Step 3: Analyze equation Q.
Q: This can be rearranged as: Here, is multiplied by a function of , which still keeps it linear because appears only to the first power. Wait carefully — but note: in the term , is simply multiplied by (independent variable), which is fine. However, the right-hand side is , still only linear in . Correction: Q is actually linear. So final: P nonlinear, Q linear. Final Answer:

Step 1: Start with the given second-order linear homogeneous differential equation:

Step 2: Form the characteristic (auxiliary) equation:

Step 3: Solve the quadratic equation:

These are purely imaginary roots.

Step 4: The general solution for roots of the form is:

Here, , so:

Step 1: Solve the differential equation. We are given the equation: This is a first-order linear differential equation of the form . The integrating factor is , where . The integrating factor is: Now multiply the entire differential equation by : The left-hand side is now the derivative of : Integrating both sides with respect to : The integral of is: So, we have: Dividing both sides by : Step 2: Apply the initial condition. At , the concentration is 200 g/L: Thus, the solution is: Step 3: Calculate the concentration at hours. Since hours = 90 minutes: Using the approximate value : Step 4: Conclusion. The brine concentration in the tank at hours is approximately:

Step 1: Solve the differential equation. We are given the equation: This is a first-order linear differential equation of the form . The integrating factor is , where . The integrating factor is: Now multiply the entire differential equation by : The left-hand side is now the derivative of : Integrating both sides with respect to : The integral of is: So, we have: Dividing both sides by : Step 2: Apply the initial condition. At , the concentration is 200 g/L: Thus, the solution is: Step 3: Calculate the concentration at hours. Since hours = 90 minutes: Using the approximate value : Step 4: Conclusion. The brine concentration in the tank at hours is approximately: