Question

The Fresnel sine integral function is defined as

$S\left(x\right)=\underset{0}{\overset{x}{\int }}\sin \left(\frac{\pi }{2}{t}^2\right)dt$

See Appendix A. Express the solution of the initial-value problem

$\frac{dy}{dx}-\left(\sin x^2\right)y=0, y\left(0\right)=5$

In terms of S(x)

Short answer

The solution of initial-value problem is;$y=5e^{\sqrt{\frac{\pi }{2}}S\left(\sqrt{\frac{2}{\pi }}x\right)}$

Step-by-step solution

Definition of an initial-value problem

The unknown function $y\left(x\right)$and its derivatives at a number $x_0$. On some interval I containing I 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 $x_0$$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.

Formula used in the equation

First order linear differential equation

$\$y\text{'}+P\left(x\right)y=f\left(x\right)\$$

The integrating factor will be $e^{\int P\left(x\right)dx}$

Determination of differential equation

First order linear differential equation

$y\text{'}+P\left(x\right)y=f\left(x\right)······\left(1\right)$and

The integrating factor will be $e^{\int P\left(x\right)dx}······\left(2\right)$

Based on the first order linear differential equation we get

$$\begin{gathered} P\left(x\right)=-\sin x^2 \\ f\left(x\right)=0 \end{gathered}$$from the given function

Now substituting above function in integrating factor formula

$e^{\int P\left(x\right)dx}=e^{-\underset{0}{\overset{x}{\int }}\sin t^{2}dt}$

Then,

$$\begin{gathered} {}^{-\underset{0}{\overset{x}{\int }}\sin t^{2}dt}=\underset{0}{\overset{x}{\int }}0dt+c \\ y{e}^{-\underset{0}{\overset{x}{\int }}\sin t^{2}dt}=c \\ y=e^{-\underset{0}{\overset{x}{\int }}\sin t^{2}dt}c······\left(3\right) \end{gathered}$$

Now, let t be

$$\begin{gathered} t=\sqrt{\frac{\pi }{2}a} \\ t=\sqrt{\frac{\pi }{2}a}\to dt=\sqrt{\frac{\pi }{2}da} \end{gathered}$$

Solving in terms of S(x)

Now solving,

$$\begin{gathered} \underset{0}{\overset{x}{\int }}\sin t^{2}dt=\underset{0}{\overset{\sqrt{\frac{2}{\pi }x}}{\int }}\sin \frac{\pi }{2}{a}^2\sqrt{\frac{\pi }{2}}da \\ =\sqrt{\frac{\pi }{2}}\underset{0}{\overset{\sqrt{\frac{2}{\pi }x}}{\int }}\sin \frac{\pi }{2}{a}^{2}da \\ =\sqrt{\frac{\pi }{2}}S\left(\sqrt{\frac{2}{\pi }}x\right)······\left(4\right) \end{gathered}$$

Let sub (4) in y eqn

$y=e^{\sqrt{\frac{\pi }{2}}S\left(\sqrt{\frac{2}{\pi }}x\right)c}·····\left(5\right)$

Now the initial conditionrole="math" localid="1663838030113" $y\left(0\right)=5$

Finding the value of c

Using the initial condition to get the value for c

$$\begin{gathered} y\left(0\right)=e^{\sqrt{\frac{\pi }{2}}S{\left(0\right)}_c} \\ 5=c \\ c=5 \end{gathered}$$

Substitution value C in equation

Now sub in the c value in eqn (5).

$$\begin{gathered} y=e^{\sqrt{\frac{\pi }{2}}S\left(\sqrt{\frac{2}{\pi }}x\right)c } \\ y=5e^{\sqrt{\frac{\pi }{2}}S\left(\sqrt{\frac{2}{\pi }}x\right)} \end{gathered}$$

Putting c = 5

Hence,the solution of initial-value problem is $y=5e^{\sqrt{\frac{\pi }{2}}S\left(\sqrt{\frac{2}{\pi }}x\right)}$