Question

(a) Show that \(\phi(t)=e^{2 t}\) is a solution of \(y^{\prime}-2 y^{\prime}-2 y=0\) and that \(y=c \phi(t)\) is also a solution of this equatio for any value of the conntanct \(y^{2}=0\) for \(t>0\), but that \(y=c \phi(t)\) is (b) Show that \(\phi(t)=1 / t\) is a solution of \(y^{\prime}+y^{2}=0\) for \(t>0\), but that \(y=c \phi(t)\) is not solution of this equation anless \(c=0\) or \(c=1 .\) Note that the equation of part (b) is nonlinear, while that of part (a) is linear.

Short answer

Question: Show that for the following differential equations, the given solutions are valid and find the general form. (a) Verify that \(y'=2y'-2y=0\) has solutions of the form \(y=c e^{2t}\), where c is any constant. (b) Verify that \(y'+y^2=0\) has a solution of the form \(y=\frac{1}{t}\) for \(t>0\), and find under which conditions there can be a solution of the form \(y=\frac{c}{t}\). Answer: (a) We have shown that \(y'=2y'-2y=0\) has a solution \(\phi(t)=e^{2t}\), and the general form of the solution is \(y=c e^{2t}\) for any constant c. (b) We have shown that \(y'+y^2=0\) has a solution \(\phi(t)=\frac{1}{t}\) for \(t>0\), and the general form of the solution is \(y=\frac{c}{t}\) with c only being 0 or 1.

Step-by-step solution

Part (a): Showing that \(\phi(t)=e^{2t}\) is a solution of the equation \(y'-2y'-2y=0\)

To show that \(\phi(t)=e^{2t}\) is a solution, we need to differentiate it with respect to \(t\) and plug it back into the equation. We have: \(\phi(t) = e^{2t}, \phi'(t) = 2e^{2t}\) Now, replace the \(y\) and its derivatives in the given differential equation by \(\phi(t)\) and \(\phi'(t)\), respectively: \(\phi'(t)-2\phi'(t)-2\phi(t) = 2e^{2t}-4e^{2t}-2e^{2t} = 0\) Since the equation holds, \(\phi(t)=e^{2t}\) is indeed a solution.

Part (a): Showing that \(y=c\phi(t)\) is a solution of the equation for any value of the constant

Now let's see if \(y=c\phi(t)\) is also a solution, where \(c\) is a constant. \(y(t) = c\phi(t) = ce^{2t}\) Now we will find the derivative: \(y'(t) = 2ce^{2t}\) Now, substitute this back into the equation: \(y'(t)-2y'(t)-2y(t) = 2ce^{2t}-4ce^{2t}-2ce^{2t} = 0\) We can observe that the equation holds for any value of the constant \(c\). Thus, \(y=c\phi(t)\) is also a solution of the given differential equation.

Part (b): Showing that \(\phi(t)=1/t\) is a solution of \(y'+y^2=0\) for \(t>0\)

To show that \(\phi(t)=1/t\) is a solution of the given equation, we need to differentiate it with respect to \(t\) and plug it back into the equation. We have: \(\phi(t) = 1/t, \phi'(t) = -1/t^2\) Now, let's plug these values into the equation: \(\phi'(t)+\phi^2(t) = -\frac{1}{t^2}+\frac{1}{t^2} = 0\) Since the equation holds for \(t>0\), \(\phi(t)=1/t\) is a solution of the given differential equation.

Part (b): Showing that \(y=c\phi(t)\) is not a solution for this equation unless \(c=0\) or \(c=1\)

Let's check if the scaling of this solution \(y(t)=c\phi(t)=c/t\) will also be a solution of this differential equation. Differentiating this expression with respect to \(t\) gives us: \(y'(t) = -c/t^2\) Now, let's plug this into the equation: \(y'(t) + y^2(t) = -\frac{c}{t^2} + \frac{c^2}{t^2} = \frac{c^2-c}{t^2}\) For this equation to hold, we only have two cases: 1. \(c^2-c=0 \Longrightarrow c(c-1)=0 \Longrightarrow c=0 \text{ or } c=1\) Thus, \(y=c\phi(t)\) is not a solution of this equation (given \(t>0\)) unless \(c=0\) or \(c=1\).