Question

Equations with the Independent Variable Missing. If a second order differential equation has the form \(y^{\prime \prime}=f(y, y)\), then the independent variable \(t\) does not appear explicitly, but only through the dependent variable \(y .\) If we let \(v=y^{\prime}\), then we obtain \(d v / d t=f(y, v)\). Since the right side of this equation depends on \(y\) and \(v\), rather than on \(t\) and \(t\), this equation is not of the form of the first order equations discussed in Chapter 2 . However, if we think of \(y\) as the independent variable, then by the chain rule \(d v / d t=(d v / d y)(d y / d t)=v(d v / d y)\). Hence the original differential equation can be written as \(v(d v / d y)=f(y, v) .\) Provided that this first order equation can be solved, we obtain \(v\) as a function of \(y .\) A relation between \(y\) and \(t\) results from solving \(d y / d t=v(y)\). Again, there are two arbitrary constants in the final result. In each of Problems 34 through 39 use this method to solve the given differential equation. $$ y y^{\prime \prime}+\left(y^{\prime}\right)^{2}=0 $$

Short answer

Question: Solve the given second-order differential equation and find the relationship between \(y\) and \(t\): \(yy^{\prime\prime} + (y^{\prime})^{2} = 0\). Answer: The relationship between \(y\) and \(t\) is given by \(y(\ln|y| - C) - y = t + k\), where \(C\) and \(k\) are constants of integration.

Step-by-step solution

Rewrite the equation in terms of \(v\) and \(y\)

Given: $$ yy^{\prime\prime} + \left(y^{\prime}\right)^{2} = 0 $$ Let \(v = y'\), so \(v' = y''\). Then, our equation becomes: $$ yv' + v^2 = 0 $$

Solve the first-order equation for \(v\)

We have: $$ y\frac{dv}{dy}=-v^2 \\ \frac{1}{v^2}\frac{dv}{dy}=-\frac{1}{y} $$ Now, we integrate both sides with respect to \(y\): $$ \int\frac{1}{v^2}\frac{dv}{dy}dy = -\int\frac{1}{y}dy \\ \int\frac{1}{v^2}dv = -\int\frac{1}{y}dy $$ The integrals give us: $$ -\frac{1}{v} = -\ln|y| + C \\ \frac{1}{v} = \ln|y| - C $$ Now, let's find \(v(y)\): $$ v(y) = \frac{1}{\ln|y| - C} $$

Find the relationship between \(y\) and \(t\)

We know that $$ \frac{dy}{dt} = v(y) = \frac{1}{\ln|y| - C} $$ Now, we separate the variables: $$ (\ln|y| - C)dy = dt $$ Integrate both sides with respect to their respective variables: $$ \int(\ln|y| - C)dy = \int dt $$ After integration, we obtain: $$ y(\ln|y| - C) - y = t + k $$ Thus, the relationship between \(y\) and \(t\) is given by: $$ y(\ln|y| - C) - y = t + k $$ where \(C\) and \(k\) are both constants of integration.