Question

Verify that the given function or functions is a solution of the differential equation. $$ 2 t^{2} y^{\prime \prime}+3 t y^{\prime}-y=0, \quad t>0 ; \quad y_{1}(t)=t^{1 / 2}, \quad y_{2}(t)=t^{-1} $$

Short answer

Question: Verify whether the functions \(y_1(t) = t^{1/2}\) and \(y_2(t) = t^{-1}\) are solutions of the given differential equation: \(2t^2y'' + 3ty' - y = 0\). Answer: After finding the first and second derivatives of each function and substituting them into the differential equation, we have confirmed that both functions satisfy the given equation. Therefore, \(y_1(t) = t^{1/2}\) and \(y_2(t) = t^{-1}\) are solutions of the differential equation \(2t^2y'' + 3ty' - y = 0\).

Step-by-step solution

Calculating the first and second derivatives of the given functions

To perform this step, we'll find the first and second derivatives of the functions \(y_1(t) = t^{1/2}\) and \(y_2(t) = t^{-1}\) For the function \(y_1(t) = t^{1/2}\): $$y_1'(t) = \frac{1}{2\sqrt{t}}$$ $$y_1''(t) = -\frac{1}{4t\sqrt{t}}$$ For the function \(y_2(t) = t^{-1}\): $$y_2'(t) = -t^{-2}$$ $$y_2''(t) = 2t^{-3}$$

Substitute the functions and their derivatives into the given differential equation

For this step, we will substitute each function and its derivatives into the given equation to verify whether it is a solution or not. For the function \(y_1(t) = t^{1/2}\): $$2t^2(-\frac{1}{4t\sqrt{t}}) + 3t(\frac{1}{2\sqrt{t}}) - t^{1/2} = -\frac{1}{2\sqrt{t}} + \frac{3\sqrt{t}}{2} - t^{1/2} = 0$$ The given differential equation holds true for \(y_1(t)\), so \(y_1(t)\) is a solution. For the function \(y_2(t) = t^{-1}\): $$2t^2(2t^{-3}) + 3t(-t^{-2}) - t^{-1} = 4t^{-1} - 3t^{-1} - t^{-1} = 0$$ The given differential equation holds true for \(y_2(t)\), so \(y_2(t)\) is a solution.

Conclusion

Since both functions, \(y_1(t) = t^{1/2}\) and \(y_2(t) = t^{-1}\), satisfy the given differential equation, we can conclude that they are indeed solutions of the equation: $$2t^2y'' + 3ty' - y = 0$$

Key concepts

Verifying Solutions

Verifying solutions to differential equations is like confirming that you have the correct key to a lock; it must fit perfectly to be considered the right match. When provided with a differential equation and a purported solution, the act of verification involves confirming that, when the solution and all of its derivatives are substituted back into the equation, the result is an identity that holds true.

For example, consider the differential equation and its proposed solutions from the exercise: $$2 t^{2} y^{\textprime \textprime}+3 t y^{\textprime}-y=0$$. Our objective is to ascertain if the functions $$y_1(t)=t^{1/2}$$ and $$y_2(t)=t^{-1}$$ are actual solutions. The verification process unfolds in a systematic fashion: compute the derivatives, substitute these, along with the original functions, into the differential equation, and simplify. If the result nullifies the equation (equals zero), the functions are indeed the keys we are looking for.

Differentiation

Differentiation is a mathematical operation that computes the rate at which a quantity changes. It's a foundational tool in calculus used to find derivatives—which tell us how fast or slow something is moving at any given point.

Consider the functions provided in our exercise, $$y_1(t) = t^{1/2}$$ and $$y_2(t) = t^{-1}$$. To verify whether these functions solve the differential equation, we need their derivatives. The first derivative, denoted $$y'(t)$$, represents the function's rate of change. As shown in the step-by-step solution, the derivatives were calculated using the power rule. In practical terms, if the functions represent physical quantities, their derivatives could be likened to velocities if the functions were positions over time.

Substitution into Differential Equations

Once we have the derivatives of our proposed solutions, the next critical step involves substitution into the original differential equation. This procedure is akin to testing a scientific theory by plugging in data to see if the predictions hold.

In our exercise, substituting the derivatives and the original functions into the differential equation allows us to check their validity as solutions. The meticulous step of substitution must be followed by an equally careful simplification of the resulting expression. If after all substitutions and simplifications, both sides of the equation are equivalent (often resulting in 0=0), the substituted function withstands the scrutiny and is confirmed as a solution to the differential equation.