Question

Finding a Particular Solution In Exercises \(35-40\) , verify that the general solution satisfies the differential equation. Then find the particular solution that satisfies the initial condition(s). $$ \begin{array}{l}{3 x^{2}+2 y^{2}=C} \\ {3 x+2 y y^{\prime}=0} \\ {y=3 \text { when } x=1}\end{array} $$

Short answer

The differential equation is verified by the general solution. The particular solution that satisfies the initial condition \(y=3\) when \(x=1\) is \(3x^2 + 2y^2 = 21\)

Step-by-step solution

Find the derivative

Using implicit differentiation on \(3x^2 + 2y^2 = C\). Differentiating both sides with respect to \(x\), we get \(6x + 4yy' = 0\). If we simplify, we obtain the value of the derivative as \(y' = -\frac{3x}{2y}\).

Verify the obtained derivative

Now, substitute \(y'\) into the given equation \(3x+2yy' = 0\). We get \(3x - 2y(\frac{3x}{2y}) = 0\), which simplifies to \(3x - 3x = 0\). Since the left side does equal to 0, as per the right side of the equation, we can confirm that the derivative we found is correct and that the general solution satisfies the differential equation.

Substitute the initial condition

To find the particular solution, we substitute the initial condition: \(y = 3\) when \(x = 1\) into the general solution. Therefore, by substituting these values into \(3x^2 + 2y^2 = C\), we get \(C = 3(1)^2 + 2(3)^2 = 3 + 18 = 21\).

Write the particular solution

Now, knowing that \(C = 21\), we can rewrite the general solution as the particular solution: \(3x^2 + 2y^2 = 21\).

Key concepts

Implicit Differentiation

Understanding implicit differentiation is essential when dealing with equations where it's not easy or possible to solve for one variable in terms of another before differentiating. In the given exercise, the equation to differentiate is not given in the form of 'y' explicitly as a function of 'x', otherwise known as an explicit function. Instead, we have an equation where 'x' and 'y' are mingled together.
Implicit differentiation involves differentiating both sides of the equation with respect to 'x' and treating 'y' as an implicit function of 'x'. In simple terms, every time you differentiate a term with 'y' in it, you multiply by 'y'' (which represents dy/dx), because you're applying the chain rule. So, when we take the derivative of the equation given in the exercise, we remember that 'y' is actually a function of 'x', and we treat it accordingly.
Here's how it applies step by step to our exercise:

• We start with the equation: \( 3x^2 + 2y^2 = C \).
• When differentiating, the term \( 3x^2 \) becomes \( 6x \) because the power rule applies.
• For the term \( 2y^2 \), we apply the chain rule. Its derivative is \( 4yy' \), because \( y^2 \) as a function of 'x' differentiates to \( 2y \), and we then multiply by \( y' \) (the derivative of 'y' with respect to 'x').
• Thus, the differentiated form becomes \( 6x + 4yy' = 0 \), which leads us to \( y' = -\frac{3x}{2y} \).

This process is crucial for solving many types of calculus problems involving related rates, optimization, and of course, differential equations.

Initial Conditions

Initial conditions provide a specific value or set of values for the variables in an equation at a particular point, usually the start of the event or process being described. These conditions allow us to find a particular solution to the differential equation from a family of general solutions.
When dealing with differential equations, we often end up with a general solution that includes a constant of integration (as represented by 'C' in our exercise). This constant represents an infinite number of possible functions that could satisfy the differential equation. To narrow this down to a single, specific function—a 'particular solution'—we use given initial conditions.
In the exercise, the initial condition provided is \( y = 3 \) when \( x = 1 \). By substituting these values into the general solution equation, we determine the value of 'C'. It is like fitting the equation to a particular point on the curve, ensuring that the solution is tailored to a specific scenario. Thus, initial conditions are fundamental to solving real-world problems, which often require a unique solution reflecting the given situation.

Verifying Solutions to Differential Equations

Verifying that a function is indeed a solution to a differential equation is as critical as finding the solution itself. Verification ensures that we've correctly applied the processes like implicit differentiation and that our general solution fits all given conditions. To verify a solution to a differential equation, we follow these steps:

• Take the proposed general solution and differentiate it as necessary according to the original differential equation. This often involves implicit differentiation if the equation is not given in explicit form.
• Plug the derivative (or derivatives, if more than one differentiation is called for) back into the original differential equation. If we end up with an identity or a true statement, our general solution is verified.
• If there are initial conditions, we substitute these into the general solution to find the constant of integration, giving us the particular solution.
• We check this particular solution by making sure it satisfies both the differential equation and the initial conditions.

In our textbook example, after we use implicit differentiation and find the value of the derivative, we confirm it by substituting it back into the differential equation. We verify that the left side of the equation equals the right side, which in this case, is zero. Furthermore, we apply the initial condition to ensure our particular solution is correct, signifying a thorough validation of our process.