Menu Close

Find-all-real-solutions-y-to-the-equation-sin-dy-dx-sin-d-2-y-dx-2-0-For-each-solution-determine-the-value-of-Q-max-r-1-n-sin-d-r-y-dx-r-giving-the-value-s-of-x-for-which-Q-ar

Tinku Tara 0 Comments
Pin




Question Number 2054 by Yozzi last updated on 01/Nov/15
Find all real solutions y to the equation sin((dy/dx))+sin((d^2 y/dx^2 ))=0 . For each solution determine the value of Q=max(Σ_(r=1) ^n sin((d^r y/dx^r ))), giving the value(s) of x for which Q arises.
$${Find}\:{all}\:{real}\:{solutions}\:{y}\:{to}\:{the}\:{equation} \\ $$$$\:\:\:\:{sin}\left(\frac{{dy}}{{dx}}\right)+{sin}\left(\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\right)=\mathrm{0}\:. \\ $$$${For}\:{each}\:{solution}\:{determine}\:{the}\:{value}\:{of} \\ $$$${Q}={max}\left(\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{sin}\left(\frac{{d}^{{r}} {y}}{{dx}^{{r}} }\right)\right),\:{giving}\:{the}\:{value}\left({s}\right) \\ $$$${of}\:{x}\:{for}\:{which}\:{Q}\:{arises}. \\ $$
Commented by 123456 last updated on 01/Nov/15
y=k
$${y}={k} \\ $$
Commented by 123456 last updated on 01/Nov/15
∣(dy/dx)∣≪1,∣(d^2 y/dx^2 )∣≪1,∣θ∣≪1⇒sin θ≈θ (dy/dx)+(d^2 y/dx^2 )=0 λ+λ^2 =λ(1+λ)=0 y=C_1 +C_2 e^(−x) −−−−−−−−−−− y=C_1 +C_2 e^(−x) (dy/dx)=−C_2 e^(−x) (d^2 y/dx^2 )=C_2 e^(−x) sin (dy/dx)+sin (d^2 y/dx^2 )=0 −sin C_2 e^(−x) +sin C_2 e^(−x) =0 (T) so y=C_1 +C_2 e^(−x) is a solution, but not sure if its the full solution.
$$\mid\frac{{dy}}{{dx}}\mid\ll\mathrm{1},\mid\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\mid\ll\mathrm{1},\mid\theta\mid\ll\mathrm{1}\Rightarrow\mathrm{sin}\:\theta\approx\theta \\ $$$$\frac{{dy}}{{dx}}+\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{0} \\ $$$$\lambda+\lambda^{\mathrm{2}} =\lambda\left(\mathrm{1}+\lambda\right)=\mathrm{0} \\ $$$${y}=\mathrm{C}_{\mathrm{1}} +\mathrm{C}_{\mathrm{2}} {e}^{−{x}} \\ $$$$−−−−−−−−−−− \\ $$$${y}=\mathrm{C}_{\mathrm{1}} +\mathrm{C}_{\mathrm{2}} {e}^{−{x}} \\ $$$$\frac{{dy}}{{dx}}=−\mathrm{C}_{\mathrm{2}} {e}^{−{x}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{C}_{\mathrm{2}} {e}^{−{x}} \\ $$$$\mathrm{sin}\:\frac{{dy}}{{dx}}+\mathrm{sin}\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{0} \\ $$$$−\mathrm{sin}\:\mathrm{C}_{\mathrm{2}} {e}^{−{x}} +\mathrm{sin}\:\mathrm{C}_{\mathrm{2}} {e}^{−{x}} =\mathrm{0}\:\left(\mathrm{T}\right) \\ $$$$\mathrm{so} \\ $$$${y}=\mathrm{C}_{\mathrm{1}} +\mathrm{C}_{\mathrm{2}} {e}^{−{x}} \\ $$$$\mathrm{is}\:\mathrm{a}\:\mathrm{solution},\:\mathrm{but}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{if}\:\mathrm{its}\:\mathrm{the} \\ $$$$\mathrm{full}\:\mathrm{solution}. \\ $$
Commented by prakash jain last updated on 01/Nov/15
Q=max(Σ_(r=1) ^n sin((d^r y/dx^r )))=Σ_(r=1) ^n sin (d^r y/dx^r ) max seems to be unnecessary function since there is only one parameter sum.
$$ \\ $$$${Q}={max}\left(\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{sin}\left(\frac{{d}^{{r}} {y}}{{dx}^{{r}} }\right)\right)=\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}\mathrm{sin}\:\frac{{d}^{{r}} {y}}{{dx}^{{r}} } \\ $$$${max}\:{seems}\:{to}\:{be}\:{unnecessary}\:{function}\:{since} \\ $$$${there}\:{is}\:{only}\:{one}\:{parameter}\:\mathrm{sum}. \\ $$
Commented by Yozzi last updated on 01/Nov/15
Sorry for the miscommunication. I was just thinking then about the value of the sum Σ_(r=1) ^n sin((d^r y/dx^r )) assumimg we use one of the solutions. So, if y=A+Be^(−x) +2nπx use it to deduce Q.
$${Sorry}\:{for}\:{the}\:{miscommunication}. \\ $$$${I}\:{was}\:{just}\:{thinking}\:{then}\:{about} \\ $$$${the}\:{value}\:{of}\:{the}\:{sum}\:\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{sin}\left(\frac{{d}^{{r}} {y}}{{dx}^{{r}} }\right)\: \\ $$$${assumimg}\:{we}\:{use}\:{one}\:{of}\:{the}\:{solutions}. \\ $$$${So},\:{if}\:{y}={A}+{Be}^{−{x}} +\mathrm{2}{n}\pi{x}\:{use}\:{it}\:{to} \\ $$$${deduce}\:{Q}. \\ $$
Commented by prakash jain last updated on 01/Nov/15
ok. understood.
$${ok}.\:{understood}. \\ $$
Answered by prakash jain last updated on 01/Nov/15
u=(dy/dx),v=(d^2 y/dx^2 ), sin u=−sin v v=2nπ−u, v=(2n+1)π+u (d^2 y/dx^2 )=(2n+1)π+(dy/dx) ....(1) y=c_1 +c_2 e^x −(2n+1)πx .....(solution A) y′=c_2 e^x −(2n+1)π, y′′=c_2 e^x ⇒sin(y′′)=−sin (y′) (d^2 y/dx^2 )=2nπ−(dy/dx) ...(2) Solution for (2) y=c_1 +c_2 e^(−x) +2nπx .....(solution B) y′=−c_2 e^(−x) +2nπ, y′′=c_2 e^(−x) ⇒sin(y′′)=−sin (y′)
$${u}=\frac{{dy}}{{dx}},{v}=\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} },\:\mathrm{sin}\:{u}=−\mathrm{sin}\:{v} \\ $$$${v}=\mathrm{2}{n}\pi−{u},\:{v}=\left(\mathrm{2}{n}+\mathrm{1}\right)\pi+{u} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\left(\mathrm{2}{n}+\mathrm{1}\right)\pi+\frac{{dy}}{{dx}}\:\:\:\:\:….\left(\mathrm{1}\right) \\ $$$${y}={c}_{\mathrm{1}} +{c}_{\mathrm{2}} {e}^{{x}} −\left(\mathrm{2}{n}+\mathrm{1}\right)\pi{x}\:\:\:\:…..\left(\mathrm{solution}\:{A}\right) \\ $$$${y}'={c}_{\mathrm{2}} {e}^{{x}} −\left(\mathrm{2}{n}+\mathrm{1}\right)\pi,\:{y}''={c}_{\mathrm{2}} {e}^{{x}} \Rightarrow\mathrm{sin}\left({y}''\right)=−\mathrm{sin}\:\left({y}'\right) \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }=\mathrm{2}{n}\pi−\frac{{dy}}{{dx}}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\left(\mathrm{2}\right) \\ $$$$\mathrm{Solution}\:\mathrm{for}\:\left(\mathrm{2}\right) \\ $$$${y}={c}_{\mathrm{1}} +{c}_{\mathrm{2}} {e}^{−{x}} +\mathrm{2}{n}\pi{x}\:\:\:\:\:\:\:\:\:\:\:…..\left(\mathrm{solution}\:\mathrm{B}\right) \\ $$$${y}'=−{c}_{\mathrm{2}} {e}^{−{x}} +\mathrm{2}{n}\pi,\:{y}''={c}_{\mathrm{2}} {e}^{−{x}} \Rightarrow\mathrm{sin}\left({y}''\right)=−\mathrm{sin}\:\left({y}'\right) \\ $$
Commented by Yozzi last updated on 01/Nov/15
Let y=A+Be^(−x) +2nπx (Solution B) y^′ =−Be^(−x) +2nπ y^(′′) =Be^(−x) y^(′′′) =−Be^(−x) y^4 =Be^(−x) .... y^n = { ((−Be^(−x) for odd n>1)),((Be^(−x) for even n≥2)) :} ∴Σ_(r=1) ^n sin(y^((r)) )=sin(2nπ−Be^(−x) )+sin(Be^(−x) )+sin(−Be^(−x) )+sin(Be^(−x) )+...+sin(y^((n)) ) Σ_(r=1) ^n sin(y^((r)) )= { ((−sinBe^(−x) odd n)),((0 even n)) :} Q=max(−sinBe^(−x) )=1 when Be^(−x) =((3π)/2)⇒e^(−x) =((3π)/(2B)) If B>0⇒−x=ln(((3π)/(2B))) x=ln(((2B)/(3π))). If B<0, let B=−T, T>0 ∴Σ_(r=1) ^n sin(y^((n)) )=sin(Te^(−x) ) Q=1⇒x=ln(((2T)/π)).
$${Let}\:{y}={A}+{Be}^{−{x}} +\mathrm{2}{n}\pi{x}\:\left({Solution}\:{B}\right) \\ $$$${y}^{'} =−{Be}^{−{x}} +\mathrm{2}{n}\pi \\ $$$${y}^{''} ={Be}^{−{x}} \\ $$$${y}^{'''} =−{Be}^{−{x}} \\ $$$${y}^{\mathrm{4}} ={Be}^{−{x}} \\ $$$$…. \\ $$$${y}^{{n}} =\begin{cases}{−{Be}^{−{x}} \:{for}\:{odd}\:{n}>\mathrm{1}}\\{{Be}^{−{x}} \:\:\:\:\:{for}\:{even}\:{n}\geqslant\mathrm{2}}\end{cases} \\ $$$$\therefore\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{sin}\left({y}^{\left({r}\right)} \right)={sin}\left(\mathrm{2}{n}\pi−{Be}^{−{x}} \right)+{sin}\left({Be}^{−{x}} \right)+{sin}\left(−{Be}^{−{x}} \right)+{sin}\left({Be}^{−{x}} \right)+…+{sin}\left({y}^{\left({n}\right)} \right) \\ $$$$\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{sin}\left({y}^{\left({r}\right)} \right)=\begin{cases}{−{sinBe}^{−{x}} \:\:{odd}\:{n}}\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{even}\:{n}}\end{cases} \\ $$$${Q}={max}\left(−{sinBe}^{−{x}} \right)=\mathrm{1}\: \\ $$$${when}\:{Be}^{−{x}} =\frac{\mathrm{3}\pi}{\mathrm{2}}\Rightarrow{e}^{−{x}} =\frac{\mathrm{3}\pi}{\mathrm{2}{B}} \\ $$$${If}\:{B}>\mathrm{0}\Rightarrow−{x}={ln}\left(\frac{\mathrm{3}\pi}{\mathrm{2}{B}}\right) \\ $$$${x}={ln}\left(\frac{\mathrm{2}{B}}{\mathrm{3}\pi}\right).\: \\ $$$${If}\:{B}<\mathrm{0},\:{let}\:{B}=−{T},\:{T}>\mathrm{0} \\ $$$$\therefore\underset{{r}=\mathrm{1}} {\overset{{n}} {\sum}}{sin}\left({y}^{\left({n}\right)} \right)={sin}\left({Te}^{−{x}} \right) \\ $$$${Q}=\mathrm{1}\Rightarrow{x}={ln}\left(\frac{\mathrm{2}{T}}{\pi}\right). \\ $$$$ \\ $$
Commented by Yozzi last updated on 01/Nov/15
y=A+Be^(−x) +2nπx ⇒y^′ =−Be^(−x) +2nπ y^(′′) =Be^(−x) ∴y^(′′) +y^′ =2nπ y=A+Be^x +(2n+1)πx y^′ =Be^x +(2n+1)π y^(′′) =Be^x y^(′′) −y^′ =−(2n+1)π
$${y}={A}+{Be}^{−{x}} +\mathrm{2}{n}\pi{x} \\ $$$$\Rightarrow{y}^{'} =−{Be}^{−{x}} +\mathrm{2}{n}\pi \\ $$$${y}^{''} ={Be}^{−{x}} \\ $$$$\therefore{y}^{''} +{y}^{'} =\mathrm{2}{n}\pi \\ $$$$ \\ $$$${y}={A}+{Be}^{{x}} +\left(\mathrm{2}{n}+\mathrm{1}\right)\pi{x} \\ $$$${y}^{'} ={Be}^{{x}} +\left(\mathrm{2}{n}+\mathrm{1}\right)\pi \\ $$$${y}^{''} ={Be}^{{x}} \\ $$$${y}^{''} −{y}^{'} =−\left(\mathrm{2}{n}+\mathrm{1}\right)\pi\: \\ $$$$ \\ $$
Commented by prakash jain last updated on 01/Nov/15
Mistake is sign of solution (A). Corrected.
$$\mathrm{Mistake}\:\mathrm{is}\:\mathrm{sign}\:\mathrm{of}\:\mathrm{solution}\:\left(\mathrm{A}\right).\:{Corrected}. \\ $$
Commented by Yozzi last updated on 02/Nov/15
Awesome.
$${Awesome}. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *