Question

In each of Problems 40 through 43 solve the given initial value problem using the methods of Problems 28 through \(39 .\) $$ y^{\prime} y^{\prime \prime}=2, \quad y(0)=1, \quad y^{\prime}(0)=2 $$

Short answer

Question: Find the function y(t) that satisfies the given second-order, nonlinear ordinary differential equation (ODE) and initial conditions: $$ y^{\prime} y^{\prime \prime} = 2, \quad y(0) = 1, \quad y^{\prime}(0) = 2 $$ Answer: The function y(t) that satisfies the given ODE and initial conditions is: $$ y(t) = \frac{1}{6}(4t + 4)^{3/2} + \frac{2}{3} $$

Step-by-step solution

Integrate the given ODE once with respect to t

Start by multiplying both sides of the equation by dt, and then integrate: $$ \int y^{\prime} y^{\prime \prime} dt = \int 2 dt \\ \int y^{\prime} d(y^{\prime}) = \int 2 dt $$ Now integrate both sides: $$ \frac{1}{2}(y^{\prime})^2 = 2t + C_1 $$

Apply the initial condition y'(0)

Use the initial condition y'(0) = 2: $$ \frac{1}{2}(2)^2 = 2(0) + C_1 \\ \Rightarrow C_1 = 2 $$ So, $$ \frac{1}{2}(y^{\prime})^2 = 2t + 2 $$

Integrate again to find y(t)

To find y(t), we must integrate the expression for y'(t). First, rearrange the equation to make it easier to integrate: $$ y^{\prime} = \sqrt{4t + 4} \\ $$ Now integrate with respect to t: $$ \int y^{\prime} dt = \int \sqrt{4t + 4} dt \\ y(t) = \int \sqrt{4t + 4} dt + C_2 $$ To find the antiderivative, use substitution. Let \(u = 4t + 4\). Then, du = 4dt. Rewrite the integral as: $$ y(t) = \frac{1}{4} \int \sqrt{u} du + C_2 \\ = \frac{1}{4} \cdot \frac{2}{3} u^{3/2} + C_2 \\ = \frac{1}{6} (4t + 4)^{3/2} + C_2 $$

Apply the initial condition y(0)

Use the initial condition y(0) = 1: $$ 1 = \frac{1}{6}(4(0) + 4)^{3/2} + C_2 \\ \Rightarrow C_2 = \frac{2}{3} $$ So the solution to the initial value problem is: $$ y(t) = \frac{1}{6}(4t + 4)^{3/2} + \frac{2}{3} $$