Question

Show that $\mathit{x}\mathbf{=}\underset{\mathbf{0}}{\overset{\mathbf{y}}{\mathbf{\int }}}\frac{\mathbf{1}}{\sqrt{{\mathbf{t}}^{\mathbf{3}}\mathbf{+}\mathbf{1}}}\mathit{d}\mathit{t}$Is an implicit solution of the initial-value problem $\mathbf{2}\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{y}}{\mathbf{d}{\mathbf{x}}^{\mathbf{2}}}\mathbf{-}\mathbf{3}{\mathit{y}}^{\mathbf{2}}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\text{'}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}$.Assume that$\mathit{y}\mathbf{⩾}\mathbf{0}$. [Hint: The integral is nonelementary. See (ii) in the Remarks at the end of section 1.1]

Short answer

Proved

Step-by-step solution

Step1:Definition of initial-value problem

The unknown function y(x) and its derivatives at a number $x_0$. On some interval I containing the problem of solving an nth-order differential equation subject to n side conditions specified at:

Solve:$\frac{d^{n}y}{d{x}^n}=f(x,y,y\text{'},...,y^{(n-1)})$

Subject to:$y\left(x_0\right)=y_0,y\text{'}\left(x_0\right)=y_1,y^{(n-1)}\left(x_0\right)=y_{(n-1)}.$

Where $y_0,y_1,...,y_{n-1}$are arbitrary constants, is called n-th order Initial Value Problem (IVP). The values of $y\left(x\right)$and its first n-1 derivatives at $y\left(x_0\right)=y_0,y\text{'}\left(x_0\right)=y_1,y^{(n-1)}\left(x_0\right)=y_{(n-1)}.$are called Initial Conditions.

Using the integration method$\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)$

Step 2: Substituting for dy/dx=v in function

Let

$$\begin{gathered} \frac{dy}{dx}=v \\ \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right) \\ \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right) \\ =\frac{d}{dx}\left(v\right) \\ =\frac{dv}{dy}\left(\frac{dy}{dx}\right) \\ =\frac{dv}{dy}v \end{gathered}$$

Step3:Substituting the above function in given equation

We assume that $y⩾0$

Now we have

$\frac{d^{2}y}{d{x}^2}=\frac{dv}{dy}v······\left(1\right)$

We have to substitute Eqn(1) in the given equation$2\frac{d^{2}y}{d{x}^2}-3y^2=0$

$$\begin{gathered} 2\frac{dv}{dy}v-3y^2=0 \\ 2\frac{dv}{dy}v=3y^2 \\ 2dv\left(v\right)=3y^{2}dy \\ 2\int vdv=3\int y^{2}dy \\ {v}^2=y^3+c \\ v=\sqrt{y^3+c} \end{gathered}$$

Now putting $v=dy/dx$

$$\begin{gathered} \frac{dy}{dx}=\sqrt{y^3+c} \\ y\text{'}=\sqrt{y^3+c} \\ (y\text{'}{)}^2=y^3+c \end{gathered}$$

Finding the value c

Now let’s, use the other 2 equations to solve the function c

$y\left(0\right)=0\text{and}y\text{'}\left(0\right)=1$

$$\begin{gathered} (y\text{'}{)}^2=y^3+c \\ {1}^2=0^3+c \\ c=1 \end{gathered}$$

Substituting the value of c and Solving

Let’s solve c=1 in equation$(y\text{'}{)}^2=y^3+c$

$$\begin{gathered} (y\text{'}{)}^2=y^3+1 \\ y\text{'}=\sqrt{y^3+c} \\ \frac{dy}{dx}=\sqrt{y^3+c} \\ \frac{dy}{\sqrt{y^3+c}}=dx \end{gathered}$$

Integrating on both sides

$$\begin{gathered} \int \frac{dy}{\sqrt{y^3+c}}=\int dx \\ \underset{0}{\overset{y}{\int }}\frac{dt}{\sqrt{t^3+1}}=x \end{gathered}$$

Hence,$\underset{0}{\overset{y}{\int }}\frac{dt}{\sqrt{t^3+1}}=x$is an implicit solution of the initial-value problem.