Question

Verify that \(y_{1}(t)=1\) and \(y_{2}(t)=t^{1 / 2}\) are solutions of the differential equation \(y y^{\prime \prime}+\) \(\left(y^{\prime}\right)^{2}=0\) for \(t>0 .\) Then show that \(c_{1}+c_{2} t^{1 / 2}\) is not, in general, a solution of this equation. Explain why this result does not contradict Theorem 3.2 .2 .

Short answer

Question: Verify that \(y_1(t) = 1\) and \(y_2(t) = t^{\frac{1}{2}}\) are solutions of the differential equation \(yy'' + (y')^2 = 0\). Show that \(c_1 + c_2t^{\frac{1}{2}}\) is not a solution, in general, and explain why this doesn't contradict Theorem 3.2.2. Answer: We have verified that \(y_1(t) = 1\) and \(y_2(t) = t^{\frac{1}{2}}\) are solutions of the given differential equation. However, \(c_1 + c_2t^{\frac{1}{2}}\) is not a solution, in general. This doesn't contradict Theorem 3.2.2 because the theorem applies only to linear homogeneous differential equations, while the given equation is non-linear.

Step-by-step solution

Compute derivatives for \(y_1(t)\)

First, we'll find the first and second derivatives of \(y_1(t)=1\). \(y_1'(t) = \frac{d(1)}{dt} = 0\) \(y_1''(t) = \frac{d(0)}{dt} = 0\)

Verify if \(y_1(t)\) is a solution for the differential equation

Now, we'll plug the values of \(y_1(t)\), \(y_1'(t)\), and \(y_1''(t)\) into the differential equation \(yy''+(y')^2=0\). \((1)(0) + (0)^2=0\) The equation is true, so \(y_1(t) = 1\) is a solution of the given differential equation.

Compute derivatives for \(y_2(t)\)

Next, we'll compute the first and second derivatives of \(y_2(t) = t^{\frac{1}{2}}\). \(y_2'(t) = \frac{1}{2}t^{-\frac{1}{2}}\) \(y_2''(t) = -\frac{1}{4}t^{-\frac{3}{2}}\)

Verify if \(y_2(t)\) is a solution for the differential equation

Now, we'll plug the values of \(y_2(t)\), \(y_2'(t)\), and \(y_2''(t)\) into the differential equation \(yy''+(y')^2=0\). \((t^{\frac{1}{2}})\left(-\frac{1}{4}t^{-\frac{3}{2}}\right)+\left(\frac{1}{2}t^{-\frac{1}{2}}\right)^2\) \(=-\frac{1}{4}+\frac{1}{4}=0\) The equation is true, so \(y_2(t) = t^{\frac{1}{2}}\) is a solution of the given differential equation.

Compute derivatives for \(c_1+c_2t^{\frac{1}{2}}\)

Now, we'll compute the first and second derivatives of \(c_1 + c_2t^{\frac{1}{2}}\). \(y'(t) = 0 + \frac{1}{2}c_2t^{-\frac{1}{2}} = \frac{1}{2}c_2t^{-\frac{1}{2}}\) \(y''(t) = -\frac{1}{4}c_2t^{-\frac{3}{2}}\)

Check if \(c_1+c_2t^{\frac{1}{2}}\) is a solution for the differential equation

Now, we'll plug the values of \(y(t)\), \(y'(t)\), and \(y''(t)\) into the differential equation \(yy''+(y')^2=0\). \(\left(c_1 + c_2t^{\frac{1}{2}}\right)\left(-\frac{1}{4}c_2t^{-\frac{3}{2}}\right)+\left(\frac{1}{2}c_2t^{-\frac{1}{2}}\right)^2\) \(=-\frac{1}{4}c_2^2t^{-\frac{1}{2}}+\frac{1}{4}c_2^2t^{-1}\neq 0\) The equation is not true, so \(c_1 + c_2t^{\frac{1}{2}}\) is not a solution of the given differential equation, in general. This result doesn't contradict Theorem 3.2.2, because the differential equation is non-linear. Theorem 3.2.2 states that the linear combination of two solutions is also a solution only for linear homogeneous differential equations. Since our given differential equation is non-linear, we cannot claim that the linear combination of the given solutions is also a solution.

Key concepts

Non-linear differential equations

In mathematics, a non-linear differential equation is one in which the dependent variable or its derivatives appear with exponents greater than one, or in combinations that are not linear. This set of equations differs greatly from linear differential equations, where the conditions are much simpler and often allow for straightforward solutions.
A characteristic of non-linear differential equations is that the principle of superposition—which states that the sum of any two solutions of a linear equation is also a solution—does not apply. This makes solving them more complex and nuanced.
In the equation provided in the exercise, each variable does not only multiply by constants. Instead, they have terms like \(yy'' + (y')^2 = 0\) which implies non-linearity due to the presence of products and powers of the derivatives. This deviation from linearity restricts the techniques and solutions applicable to finding their solutions.

Solution verification

Verifying a solution involves plugging the proposed solution back into the original differential equation to check if it satisfies the equation.
In this exercise, we checked two potential solutions, \(y_1(t) = 1\) and \(y_2(t) = t^{1/2}\). For both, the substituents proved satisfactory, as they turned the equation \(yy'' + (y')^2 = 0\) into a true statement. Breaking this process into manageable steps ensures accuracy and confidence in our solutions.
It's crucial to verify because these solutions were derived under the assumption they satisfy the given differential equation; however, there are cases, especially in non-linear equations, where unexpected results or additional considerations can arise. This is evident in the third case analyzed, \(c_1 + c_2 t^{1/2}\), where calculation showed the result did not hold as a solution in a general sense. This highlights that additional assumptions or restrictions might apply.

Theorem 3.2.2

Theorem 3.2.2 is a fundamental result usually dealing with solutions to linear homogeneous differential equations. It states that if \(y_1(t)\) and \(y_2(t)\) are solutions, then any linear combination of them \(c_1y_1(t) + c_2y_2(t)\) is also a solution. However, a crucial caveat in this theorem is its limitation to linear equations.
In our exercise, the differential equation is non-linear. Therefore, while \(y_1(t)\) and \(y_2(t)\) independently satisfied the equation, their linear combination did not necessarily result in a solution. This non-linearity means the principle highlighted in Theorem 3.2.2 does not apply.
The pivotal point here is recognizing the difference in behavior between linear and non-linear systems. For students, grasping this distinction can help avoid mistakes in wrongly applying the superposition principle to non-linear equations.