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If $f: \mathbb{R}\rightarrow \mathbb{R}: (t,y) \mapsto f(t,y)$ is a differentiable function and $y:\mathbb{R}\rightarrow \mathbb{R}$ satisfies the differential equation:

$y'(t)=f(t,y(t))$

Find $y''$ and $y'''$.


Has anyone an idea how to solve such problems?

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closed as off-topic by Namaste, zhoraster, Davide Giraudo, C. Falcon, Leucippus Jan 27 '17 at 0:18

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You have to use the "chain rule".

I will change slightly the notation for the sake of clarity :

Suppose that $\forall t\in\mathbb{R},\,u'(t)=f(t,u(t))$ where $f:\mathbb{R}\to\mathbb{R},\,(x,y)\mapsto f(x,y)$

Then you get :

$$u''(t)=\frac{d}{dt}(t)\,\frac{\partial f}{\partial x}(t,u(t))+\frac{d}{dt}(u(t))\,\frac{\partial f}{\partial y}(t,u(t))$$

or, simply stated :

$$u''(t)=\frac{\partial f}{\partial x}(t,u(t))+u'(t)\,\frac{\partial f}{\partial y}(t,u(t))=\frac{\partial f}{\partial x}(t,u(t))+f(t,u(t))\,\frac{\partial f}{\partial y}(t,u(t))$$

And, after differentiating twice :

$$u'''(t)=\left[\frac{\partial^2f}{\partial x^2}(t,u(t))+u'(t)\frac{\partial^2f}{\partial x\partial y}(t,u(t))\right]+\frac{d}{dt}\left[f(t,u(t))\right]\frac{\partial f}{\partial y}(t,u(t))+f(t,u(t))\left[\frac{\partial^2f}{\partial x\partial y}(t,u(t))+u'(t)\frac{\partial^2f}{\partial y^2}(t,u(t))\right]$$

And finally :

$$u'''(t)=\frac{\partial^2f}{\partial x^2}(t,u(t))+f(t,u(t))\frac{\partial^2f}{\partial x\partial y}(t,u(t))+\left[\frac{\partial f}{\partial x}(t,u(t))+f(t,u(t))\,\frac{\partial f}{\partial y}(t,u(t))\right]\frac{\partial f}{\partial y}(t,u(t))+f(t,u(t))\left[\frac{\partial^2f}{\partial x\partial y}(t,u(t))+f(t,u(t)\frac{\partial^2f}{\partial y^2}(t,u(t))\right]$$

which can be reduced, in standard notation, to :

$$\boxed{u''(t)=f_x+ff_y}$$

and

$$\boxed{u'''(t)=f_{xx}+2ff_{xy}+f_xf_y+ff_y^2+f^2f_{xy}}$$

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Hint. Using the chain rule, we get $$ y''(t) = \partial_t \def\f#1{f\bigl(#1\bigr)}\f{t,y(t)} + \partial_y \f{t,y(t)}y'(t) = \partial_t\f{t,y(t)} + \partial_y \f{t,y(t)}\f{t,y(t)}. $$ To compute $y'''$ just differentiate again and replace $y'(t)$ by $\f{t,y(t)}$, as we did above.

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    $\begingroup$ If you write $\partial_y \f{t,y(t)}\f{t,y(t)}$ it's difficult to know what is to be differentiated with respect to $y$. Instead, try $\f{t,y(t)}\partial_y \f{t,y(t)}$ or $\left(\partial_y \f{t,y(t)}\right)\f{t,y(t)}$ $\endgroup$ – Arthur Jan 22 '17 at 10:56
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after the chain rule we get $$y''=f_t+f_y\cdot y'$$ with $$y'=f$$ we get for the second derivative $$y''=f_t+f_yf$$ further we get $$y'''=f_{tt}+f_{ty}y'+f_{yy}y'^2+f_{yt}y'+f_yy''$$ substituting $$y'=f$$ we get $$y'''=f_{tt}+f_{ty}f+f_{yy}f+f_{yy}f^2+f_{yt}f+f_y(f_t+f_yf)$$

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    $\begingroup$ I would suggest substituting $y'$ with $f$. $\endgroup$ – Arthur Jan 22 '17 at 10:55
  • $\begingroup$ Hi Thanks, can you please elaborate how you get the first expression from the chain (and product?) rule. Thx $\endgroup$ – Averroes2 Jan 22 '17 at 12:03
  • $\begingroup$ we have $$y'=f(t,y)$$ and $$y=y(t)$$ thus we get $$y''=f_t+f_y\cdot y'$$ since we have $$y=y(t)$$ using the chain rule $\endgroup$ – Dr. Sonnhard Graubner Jan 22 '17 at 12:06

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