Question

Can \(y=\sin \left(t^{2}\right)\) be a solution on an interval containing \(t=0\) of an equation \(y^{\prime \prime}+p(t) y^{\prime}+\) \(q(t) y=0\) with continuous coefficients? Explain your answer.

Short answer

Based on the step-by-step solution, we find that at \(t = 0\), the equation \(2 + 0 = 0\) is false. Therefore, the function \(y=\sin(t^2)\) cannot be a solution on an interval containing \(t=0\) for the given equation \(y^{\prime\prime}+p(t)y^{\prime}+q(t)y=0\).

Step-by-step solution

Find the first derivative of y

To find the first derivative, apply the chain rule: \(y'(t) = \frac{dy}{dt} = \frac{d}{dt}\sin(t^2) = (\cos(t^2))\cdot(2t)\), So, the first derivative is \(y'(t) = 2t\cos(t^2)\).

Find the second derivative of y

Now, let's find the second derivative by differentiating \(y'(t)\): \(y''(t) = \frac{d^2y}{dt^2} = \frac{d}{dt}(2t\cos(t^2))\). Applying the product rule and the chain rule, we get:; \(y''(t) = 2\cos(t^2) - 4t^2\sin(t^2)\).

Substitute the derivatives and the original function into the given equation

Now we substitute \(y=\sin(t^2)\), \(y'=2t\cos(t^2),\) and \(y''=2\cos(t^2)-4t^2\sin(t^2)\) into the given equation: \((2\cos(t^2)-4t^2\sin(t^2)) + p(t)(2t\cos(t^2)) + q(t)\sin(t^2) = 0\) We are trying to find if there exists \(p(t)\) and \(q(t)\) such that this equation is satisfied.

Verify the equation at t = 0

Let's consider the equation at the point \(t = 0\): \((2\cos(0) - 4(0)^2\sin(0)) + p(0)(2(0)\cos(0)) + q(0)\sin(0) = 0\) Simplifying it, we get: \(2 + 0 = 0\). This equation is false. So, it is impossible for \(y=\sin(t^2)\) to be a solution on an interval containing \(t=0\) for the given equation with continuous coefficients as it does not satisfy the equation at \(t = 0\).

Key concepts

Chain Rule

The chain rule is an essential tool in calculus, especially when dealing with composite functions. A composite function is when you have a function inside another function. The chain rule helps us differentiate these types of functions by requiring us to multiply the derivative of the outer function by the derivative of the inner function.
For example, consider the function \(y = \sin(t^2)\). This is a composite function because it's a sine function with an input of \(t^2\), an inner function. To find \(y'(t)\), we first differentiate \(\sin(t^2)\) with respect to its argument \(t^2\). The derivative of \(\sin\) is \(\cos\), so the derivative of \(\sin(t^2)\) is \(\cos(t^2)\).
Then, according to the chain rule, we multiply by the derivative of \(t^2\), which is \(2t\). This leads us to:

• \(y'(t) = 2t\cos(t^2)\)

As such, the chain rule allows us to break down the differentiation process into manageable steps, making it easier to tackle complex expressions.

Second Derivative

The second derivative is essentially the derivative of the first derivative. It provides information about the curvature or concavity of the original function, \(y\). If \(y''(t) > 0\), the function is concave up, while if \(y''(t) < 0\), it is concave down. This concept is instrumental in understanding the behavior of curves and graphs.
If we take the function \(y = \sin(t^2)\), we have already determined its first derivative is \(y'(t) = 2t\cos(t^2)\). To find the second derivative, \(y''(t)\), we need to differentiate this result again. Here, both the product rule and the chain rule come into play, since \(2t\cos(t^2)\) is the product of two functions: \(2t\) and \(\cos(t^2)\).
Applying the product rule:

• The derivative of \(2t\) is \(2\).
• The derivative of \(\cos(t^2)\) is \(-2t\sin(t^2)\), using the chain rule.

Putting these together, we get:

• \(y''(t) = 2\cos(t^2) - 4t^2\sin(t^2)\)

This second derivative is crucial for solving differential equations, as it allows us to evaluate the rate at which the rate of change (the first derivative) is changing.

Product Rule

When dealing with the differentiation of products of functions, the product rule is an invaluable tool. The product rule states that the derivative of a product of two functions is the first function times the derivative of the second plus the second function times the derivative of the first.
Mathematically, if \(u(t)\) and \(v(t)\) are functions of \(t\), then the derivative \((uv)'\) is:

• \((uv)' = u'v + uv'\)

In our example with \(y = \sin(t^2)\), when finding \(y''(t)\), we employed the product rule on the expression \(y'(t) = 2t\cos(t^2)\), as it's a product of \(2t\) and \(\cos(t^2)\).
Here's how it was applied:

• \(u = 2t\) with \(u' = 2\)
• \(v = \cos(t^2)\) with \(v' = -2t\sin(t^2)\) (using the chain rule)

Applying the product rule, we get:
\(y''(t) = 2t(-2t\sin(t^2)) + 2\cos(t^2) = -4t^2\sin(t^2) + 2\cos(t^2)\)
The product rule, as demonstrated, simplifies the differentiation of products, which is vital in handling equations involving multiple variable functions.

Continuous Coefficients

In differential equations, coefficients are often functions of the independent variable, such as \(t\) in our example. Coefficients are deemed continuous if they have no breaks, jumps, or gaps in their values across the interval being considered. This continuity is crucial when solving differential equations, as it ensures the solutions behave predictably and smoothly.
In the given differential equation format \(y'' + p(t) y' + q(t) y = 0\), both \(p(t)\) and \(q(t)\) are coefficients that need to be continuous over the interval for standard solution techniques to apply. Continuous coefficients make it possible to apply methods like the Picard-Lindelöf theorem to assure the existence and uniqueness of a solution.
In our exercise, we needed to identify if \(y = \sin(t^2)\) could satisfy this equation with continuous coefficients; however, upon evaluating the expression at \(t = 0\), it did not satisfy the equation, indicating a solution in this case cannot exist under the constraint of continuous coefficients.
Continuous coefficients are an important consideration in differential equations to ensure that solutions exist and that they smoothly depend on the initial conditions and the form of the differential equation within a given interval.