Question

Verify that the given function is a solution of the differential equation that follows it. $$u(t)=C e^{1 /\left(4 r^{4}\right)} ; u^{\prime}(t)+\frac{1}{t^{5}} u(t)=0$$

Short answer

Question: Verify if the function $$u(t) = C e^{1 /(4 t^{4})}$$ is a solution to the differential equation $$u^{\prime}(t) + \frac{1}{t^{5}} u(t) = 0$$. Answer: The given function $$u(t) = C e^{1 /(4 t^{4})}$$ is NOT a solution to the differential equation $$u^{\prime}(t) + \frac{1}{t^{5}} u(t) = 0$$.

Step-by-step solution

Find the derivative of u(t)

First, let's find the derivative of the given function $$u(t) = C e^{1 /(4 t^{4})}$$. By applying the chain rule, we get: $$u^{\prime}(t) = C e^{1 /(4 t^{4})} \cdot \frac{d (1/(4 t^{4}))}{dt}$$ Now, we need to find the derivative of $$1/(4 t^{4})$$ which can be written as $$(4 t^{4})^{-1}$$. Using the power rule, we get: $$\frac{d (1/(4 t^{4}))}{dt} = -1 \cdot (4 t^{4})^{-2} \cdot \frac{d (4t^{4})}{dt}$$ Now find the derivative of $$4t^{4}$$, which is $$16t^3$$. Combining everything, we have: $$u^{\prime}(t) = -C e^{1 /(4 t^{4})} \cdot (4 t^{4})^{-2} \cdot 16t^3$$

Verify the differential equation

Now that we have found u'(t), we need to verify if the differential equation holds true for both u(t) and u'(t). The given differential equation is: $$u^{\prime}(t) + \frac{1}{t^{5}} u(t) = 0$$ Substituting u(t) and u'(t) into the differential equation, we have: $$(-C e^{1 /(4 t^{4})} \cdot (4 t^{4})^{-2} \cdot 16t^3) + \frac{1}{t^{5}}(C e^{1 /(4 t^{4})}) = 0$$ Now, we see that both terms have a common factor of $$C e^{1 /(4 t^{4})}$$, which we can factor out: $$C e^{1 /(4 t^{4})} [(- (4 t^{4})^{-2} \cdot 16t^3) + \frac{1}{t^{5}}] = 0$$ Since $$C e^{1 /(4 t^{4})}$$ can't be zero, the only way for the equation to hold true is if the term inside the brackets is equal to zero: $$(- (4 t^{4})^{-2} \cdot 16t^3) + \frac{1}{t^{5}} = 0$$ Now, let's simplify this equation: $$(- 16t^3 (4 t^{4})^{-2} ) + \frac{1}{t^{5}} = 0$$ Multiply both sides with $$t^{5}$$: $$- 16t^3 (4 t^{4})^{-2} (t^{5}) +1 = 0$$ Now notice that $$(4t^4)^{-2} t^5 = \frac{t^5}{(4t^4)^{2}} = \frac{1}{t^3}$$, and hence we get: $$-16t^3 \cdot \frac{1}{t^3} +1 = 0$$ Which simplifies to: $$-16 +1 = 0$$ However, this equation is not true. Therefore, the given function u(t) is NOT a solution to the differential equation.

Key concepts

Chain Rule

The chain rule is a fundamental concept in calculus that allows us to find the derivative of a composite function. In simpler terms, if a function is made up of two or more smaller functions, the chain rule helps us differentiate it effectively.
When using the chain rule, you determine the derivative of the outer function and multiply it by the derivative of the inner function.

• For example, for a composite function \(f(g(x))\), the derivative can be found using the formula \(f'(g(x)) \cdot g'(x)\).
• In the original solution, the chain rule is used to differentiate \(u(t) = C e^{1 /(4 t^{4})}\). The function \(e^{1 /(4 t^{4})}\) is the outer function and \(1 /(4 t^{4})\) is the inner function.

The chain rule is incredibly powerful and necessary for differentiating complex functions in calculus.

Derivative

Differentiation is the process of finding a derivative, which represents how a function changes as its input changes. Derivatives are foundational in calculus, capturing the rate of change or the slope of a function at any given point.
To find a derivative, you must follow specific rules or theorems, such as the power rule or chain rule. Differentiation can be applied to various types of functions, and is used to determine velocity, acceleration, and optimize functions in calculus.

• For the function \(u(t) = C e^{1 /(4 t^{4})}\), the derivative \(u'(t)\) was found using both the power rule and the chain rule.
• This involves taking the derivative of the exponential function and multiplying it by the derivative of its exponent.

Understanding how to compute derivatives is crucial when solving problems involving rates of change in mathematics.

Power Rule

The power rule is one of the simplest rules in differentiation and is used when taking the derivative of a term raised to a power. According to the power rule, if you have a term \(x^n\), its derivative is \(n \cdot x^{n-1}\).
This rule makes it straightforward to handle polynomials and terms with simple exponents.

• In the example, the power rule is applied while differentiating \((4 t^4)^{-1}\), part of the exponent in the function \(u(t)\).
• The derivative of \(4t^4\) simplifies to \(16t^3\) when using this rule.

Knowing the power rule allows you to quickly differentiate terms where power terms are involved, especially when combined with other rules like the chain rule.

Verification of Solutions

Verification of solutions involves checking whether a proposed solution actually satisfies a given equation, in this case, a differential equation. This step is crucial to ensure the correctness of a solution.
In the process, you substitute the function and its derivative back into the original equation to see if it holds true for all variable values.

• For the differential equation \(u'(t) + \frac{1}{t^5} u(t) = 0\), you would plug in both \(u(t)\) and \(u'(t)\) and simplify.
• If the equation balances or simplifies to zero, then the function is indeed a solution. Otherwise, it is not.

In the original exercise, although the calculations seemed correct, the given function was ultimately shown not to meet the differential equation's requirements, demonstrating the importance of verification.