Question

Equations with the Dependent Variable Missing. For a scond order differential equation of the form \(y^{\prime \prime}=f\left(t, y^{\prime}\right),\) the substitution \(v=y^{\prime}, v^{\prime}=y^{\prime \prime}\) leads to a first order equation of the form \(v^{\prime}=f(t, v)\). If this equation can be solved for \(v\), then \(y\) can be obtained by integrating \(d y / d t=v .\) Note that one arbitrary constant is obtained in solving the first order equation for \(v,\) and a second is introduced in the integration for \(y\). In each of Problems 28 through 33 use this substitution to solve the given equation. $$ t^{2} y^{\prime \prime}+2 t y^{\prime}-1=0, \quad t>0 $$

Short answer

Answer: The general solution of the given second-order differential equation is \(y(t) = \ln{t} - \frac{1}{3}t^3 + C_1t + C_2\), where \(C_1\) and \(C_2\) are constants of integration.

Step-by-step solution

Perform the substitution

Substitute \(v=y'\) and \(v'=y''\) into the given equation. $$ t^{2}v'+2tv-1=0 $$

Solve for \(v'\)

Isolate \(v'\) in the equation: $$ v' = \frac{1-2tv}{t^2} $$

Integrate the equation to find \(v\)

Integrate the equation for \(v'\) with respect to \(t\): $$ v(t) = \int \frac{1-2tv}{t^2} dt $$ Using the power rule for integration, we find: $$ v(t) = \int \frac{1}{t^2} dt - 2\int t dt = -\frac{1}{t} - t^2 + C_1 $$ where \(C_1\) is the constant of integration.

Substitute back for \(y'\)

Now that we have found \(v(t)\), substitute back \(y'\) for \(v\). Recall that \(v=y'\): $$ y' = -\frac{1}{t} - t^2 + C_1 $$

Integrate to find \(y\)

Integrate the equation for \(y'\) with respect to \(t\) to find \(y(t)\): $$ y(t) = \int (-\frac{1}{t} - t^2 + C_1) dt $$ Using the power rule and factoring out the constant, we find: $$ y(t) = \ln{t} - \frac{1}{3}t^3 + C_1t + C_2 $$ where \(C_2\) is another constant of integration. So, the solution to the given second-order differential equation is: $$ y(t) = \ln{t} - \frac{1}{3}t^3 + C_1t + C_2 $$