I have encountered with a problem, and I can't follow the intermediate steps. Consider the following differential equation:
$$ y^{\prime \prime} = -2y + f(y)-0.5y^3 $$
Here, the first derivative is expressed as the following integral:
$$ (y^{\prime})^2 = 2\int_0^{y^\prime} y^\prime dy^\prime $$
Without giving the details of the intermediate steps, this equation is converted into the following equation using the differential equation given above:
$$ (y^{\prime})^2 = -2 \int_{Y}^{y} \left( 2y - f(y)+0.5y^3 \right) dy$$
where $Y$ is the maximum value of $y$ when $y^\prime$ is zero. Here, I would like to learn how to express differential of a derivative $ dy^\prime $ in terms of $dy$, and end up with the above equation. If I am not wrong, the following equality must hold:
$$ y^\prime dy^\prime = y^{\prime \prime} dy $$
How can I prove this equality? I am confused to think a differential like $ dy^\prime = d(\frac{dy}{dx}) $. Any help would be appreciated. Thanks.
$$y^\prime dy^\prime = y^{\prime \prime} dy$$ Simply because: $$y''dy=\left (\dfrac {dy'}{dx} \right )dy=y'dy'$$
Note that $$y^{\prime \prime} = -2y + f(y)-0.5y^3$$ Multiply by $2y'$ $$2y'y^{\prime \prime} = -4yy' + 2y'f(y)-y^3y'$$ $$(y'^2)'= -4yy' + 2y'f(y)-y^3y'$$ And integrate since $2y''y'=(y'^2)'$ $$\int (y'^2)'dx=\int ( -4yy' + 2y'f(y)-y^3y')dx$$ But since $y'dx=dy$ we simplify the integrand: $$y'^2=\int ( -4y + 2f(y)-y^3)dy$$ $$y'^2=-2\int ( 2y -f(y)+\dfrac 12 y^3)dy$$