Turning an initial value problem into an integral equation Show that the initial value problem(IVP)
$$y'' + y^2 − 1 = 0, \qquad y(0) = 0, \qquad y'(0) = 1$$
is equivalent to the integral equation
$$y(t) = t − \int_{0}^{t} \int_{0}^{T}(y^2(s) − 1)dsdT$$
 A: The initial value problem (IVP)
$$y''=1-y^2, \quad y(0)=y_0, \quad y'(0)=y_1$$
is equivalent with the first order IVP
$$\left.\eqalign{y'&=v\cr
v'&=1-y^2\cr}\right\}\ ,\quad y(0)=y_0,\quad v(0)=y_1\ ,$$
and this in turn can be rewritten as
$$\left.\eqalign{y(t)&=y_0+\int_0^t v(\tau)\>d\tau \cr
v(t)&=y_1+\int_0^t\bigl(1-y^2(s)\bigr)\>ds \cr}\right\}\ .\tag{1}$$
Plugging $v(t)$ from the second formula $(1)$ into the first fromula one obtains
$$y(t)=y_0+y_1 t+\int_0^t \int_0^\tau\bigl(1-y^2(s)\bigr)\>ds\>d\tau\ .\tag{2}$$
An easy interchange of the order of integration shows that for any function $s\mapsto g(s)$ one has
$$\int_0^t \int_0^\tau g(s)\>ds\>d\tau=\int_0^t (t-s)g(s)\>ds\ .$$
Therefore $(2)$ is equivalent to
$$y(t)=y_0+y_1 t+\int_0^t (t-s)\bigl(1-y^2(s)\bigr)\>ds\ .\tag{3}$$
Conversely, if $t\mapsto y(t)$ satisfies $(3)$ it satisfies $(2)$. Defining $t\mapsto v(t)$ by the second formula $(1)$ then shows that $(1)$ holds, which is equivalent to the original IVP.
A: HINT:
$$y''(x)+y(x)^2-1=0\Longleftrightarrow$$
$$y''(x)=1-y(x)^2\Longleftrightarrow$$
$$y''(x)y(x)=y(x)\left(1-y(x)^2\right)\Longleftrightarrow$$
$$\int y''(x)y(x)\space\text{d}x=\int y(x)\left(1-y(x)^2\right)\space\text{d}x\Longleftrightarrow$$
$$\frac{y'(x)^2}{2}=\text{C}-\frac{y(x)^3}{3}+y(x)\Longleftrightarrow$$
$$y'(x)^2=2\left(y(x)-\frac{y(x)^3}{3}\right)+2\text{C}\Longleftrightarrow$$
$$y'(x)=\pm\sqrt{2\left(y(x)-\frac{y(x)^3}{3}\right)+2\text{C}}\Longleftrightarrow$$
$$\frac{y'(x)}{\sqrt{2\left(y(x)-\frac{y(x)^3}{3}\right)+2\text{C}}}=\pm1\Longleftrightarrow$$
$$\int\frac{y'(x)}{\sqrt{2\left(y(x)-\frac{y(x)^3}{3}\right)+2\text{C}}}\space\text{d}x=\pm\int 1\space\text{d}x$$
