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Question Number 59361 by Andrew Foxman last updated on 08/May/19
Pls help r^2 r′′=C where r(t) is a function and C is a constant
$${Pls}\:{help} \\ $$$${r}^{\mathrm{2}} {r}''={C}\:{where}\:{r}\left({t}\right)\:{is}\:{a}\:{function}\:{and} \\ $$$${C}\:{is}\:{a}\:{constant} \\ $$
Commented by kaivan.ahmadi last updated on 09/May/19
(d^2 r/dt^2 )=(C/r^2 )⇒(dr/dt)=∫(C/r^2 )dt=(C/r^2 )t+C_1 ⇒ r=((Ct^2 )/r^2 )+C_1 t+C_2 ⇒Ct^2 +(C_1 t+C_2 )r^2 −r^3 =0
$$\frac{{d}^{\mathrm{2}} {r}}{{dt}^{\mathrm{2}} }=\frac{{C}}{{r}^{\mathrm{2}} }\Rightarrow\frac{{dr}}{{dt}}=\int\frac{{C}}{{r}^{\mathrm{2}} }{dt}=\frac{{C}}{{r}^{\mathrm{2}} }{t}+{C}_{\mathrm{1}} \Rightarrow \\ $$$${r}=\frac{{Ct}^{\mathrm{2}} }{{r}^{\mathrm{2}} }+{C}_{\mathrm{1}} {t}+{C}_{\mathrm{2}} \Rightarrow{Ct}^{\mathrm{2}} +\left({C}_{\mathrm{1}} {t}+{C}_{\mathrm{2}} \right){r}^{\mathrm{2}} −{r}^{\mathrm{3}} =\mathrm{0} \\ $$
Commented by mr W last updated on 09/May/19
it is wrong sir! r in ∫(C/r^2 )dt is a function of t, i.e. r=r(t), you can not treat r(t) as a constant, ∫(C/r^2 )dt≠(C/r^2 )t+C_1
$${it}\:{is}\:{wrong}\:{sir}! \\ $$$${r}\:{in}\:\int\frac{{C}}{{r}^{\mathrm{2}} }{dt}\:{is}\:{a}\:{function}\:{of}\:{t},\:{i}.{e}.\:{r}={r}\left({t}\right), \\ $$$${you}\:{can}\:{not}\:{treat}\:{r}\left({t}\right)\:{as}\:{a}\:{constant}, \\ $$$$\int\frac{{C}}{{r}^{\mathrm{2}} }{dt}\neq\frac{{C}}{{r}^{\mathrm{2}} }{t}+{C}_{\mathrm{1}} \\ $$
Commented by maxmathsup by imad last updated on 09/May/19
r is a function of variable t....
$${r}\:{is}\:{a}\:{function}\:{of}\:{variable}\:{t}…. \\ $$
Commented by Andrew Foxman last updated on 09/May/19
The mistery of this equation remains... Anyone can give their answer now and here... If you are a hero in math resolve this equation...
$${The}\:{mistery}\:{of}\:{this}\:{equation}\:{remains}… \\ $$$${Anyone}\:{can}\:{give}\:{their}\:{answer}\:{now}\:{and}\:{here}… \\ $$$${If}\:{you}\:{are}\:{a}\:{hero}\:{in}\:{math}\:{resolve}\:{this}\:{equation}… \\ $$
Commented by MJS last updated on 10/May/19
I think r(t) can only be given implicitly
$$\mathrm{I}\:\mathrm{think}\:{r}\left({t}\right)\:\mathrm{can}\:\mathrm{only}\:\mathrm{be}\:\mathrm{given}\:\mathrm{implicitly} \\ $$
Commented by MJS last updated on 10/May/19
this is the answer from a friend who owns a calculator. I cannot explain it, it′s just the output of that machine: ((Cln (1/(r(t))))/( (√B^3 )))+((r(t)(√((Br(t)−2C)/(r(t)))))/B)−((2Cln ((√((Br(t)−2C)/(r(t))))−(√B)))/( (√B^3 )))=t+A with A, B, C ∈R it can be further simplyfied when r(t)>0 ∧ b>0 2Cln((√(Br(t)−2C))−(√(Br(t))))−(√(Br(t))Br(t)−2C))=−(√B^3 )(t+A)
$$\mathrm{this}\:\mathrm{is}\:\mathrm{the}\:\mathrm{answer}\:\mathrm{from}\:\mathrm{a}\:\mathrm{friend}\:\mathrm{who}\:\mathrm{owns}\:\mathrm{a} \\ $$$$\mathrm{calculator}.\:\mathrm{I}\:\mathrm{cannot}\:\mathrm{explain}\:\mathrm{it},\:\mathrm{it}'\mathrm{s}\:\mathrm{just}\:\mathrm{the} \\ $$$$\mathrm{output}\:\mathrm{of}\:\mathrm{that}\:\mathrm{machine}: \\ $$$$\frac{{C}\mathrm{ln}\:\frac{\mathrm{1}}{{r}\left({t}\right)}}{\:\sqrt{{B}^{\mathrm{3}} }}+\frac{{r}\left({t}\right)\sqrt{\frac{{Br}\left({t}\right)−\mathrm{2}{C}}{{r}\left({t}\right)}}}{{B}}−\frac{\mathrm{2}{C}\mathrm{ln}\:\left(\sqrt{\frac{{Br}\left({t}\right)−\mathrm{2}{C}}{{r}\left({t}\right)}}−\sqrt{{B}}\right)}{\:\sqrt{{B}^{\mathrm{3}} }}={t}+{A} \\ $$$$\mathrm{with}\:{A},\:{B},\:{C}\:\in\mathbb{R} \\ $$$$\mathrm{it}\:\mathrm{can}\:\mathrm{be}\:\mathrm{further}\:\mathrm{simplyfied}\:\mathrm{when}\:{r}\left({t}\right)>\mathrm{0}\:\wedge\:{b}>\mathrm{0} \\ $$$$\mathrm{2}{C}\mathrm{ln}\left(\sqrt{{Br}\left({t}\right)−\mathrm{2}{C}}−\sqrt{{Br}\left({t}\right)}\right)−\sqrt{\left.{Br}\left({t}\right)\right){Br}\left({t}\right)−\mathrm{2}{C}}=−\sqrt{{B}^{\mathrm{3}} }\left({t}+{A}\right) \\ $$
Answered by alex041103 last updated on 10/May/19
Let′s try r(t)=−At^α ⇒r^2 =A^2 t^(2α) ⇒r′′=−Aα(α−1)t^(α−2) ⇒r^2 r′′=−A^3 α(α−1)t^(3α−2) =C ⇒ { ((3α−2=0)),((A^3 α(1−α)=C)) :} ⇒α=(2/3) ⇒A^3 (2/3)(1−(2/3))=(2/9)A^3 =C⇒A^3 =(9/2)C ⇒A=(((9C)/2))^(1/3) ⇒One solution is r(t)=−(((9C t^2 )/2))^(1/3)
$${Let}'{s}\:{try}\:{r}\left({t}\right)=−{At}^{\alpha} \\ $$$$\Rightarrow{r}^{\mathrm{2}} ={A}^{\mathrm{2}} {t}^{\mathrm{2}\alpha} \\ $$$$\Rightarrow{r}''=−{A}\alpha\left(\alpha−\mathrm{1}\right){t}^{\alpha−\mathrm{2}} \\ $$$$\Rightarrow{r}^{\mathrm{2}} {r}''=−{A}^{\mathrm{3}} \alpha\left(\alpha−\mathrm{1}\right){t}^{\mathrm{3}\alpha−\mathrm{2}} ={C} \\ $$$$\Rightarrow\begin{cases}{\mathrm{3}\alpha−\mathrm{2}=\mathrm{0}}\\{{A}^{\mathrm{3}} \alpha\left(\mathrm{1}−\alpha\right)={C}}\end{cases} \\ $$$$\Rightarrow\alpha=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$$\Rightarrow{A}^{\mathrm{3}} \frac{\mathrm{2}}{\mathrm{3}}\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{3}}\right)=\frac{\mathrm{2}}{\mathrm{9}}{A}^{\mathrm{3}} ={C}\Rightarrow{A}^{\mathrm{3}} =\frac{\mathrm{9}}{\mathrm{2}}{C} \\ $$$$\Rightarrow{A}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{9}{C}}{\mathrm{2}}} \\ $$$$\Rightarrow{One}\:{solution}\:{is}\:{r}\left({t}\right)=−\sqrt[{\mathrm{3}}]{\frac{\mathrm{9}{C}\:{t}^{\mathrm{2}} \:}{\mathrm{2}}} \\ $$$$ \\ $$
Answered by aleks041103 last updated on 07/Sep/20
Observe: r′′=(d/dt)((dr/dt))=(dv/dt)=(dv/dr) (dr/dt)=v(dv/dr) , where v=(dr/dt) Then: vdv=((Cdr)/r^2 )=−Cd((1/r))=−d((C/r)) vdv=(1/2)d(v^2 )=d((v^2 /2)) ⇒d((v^2 /2))=−d((C/r)) d((v^2 /2) + (C/r))=0⇒(v^2 /2)=(A/2)−(C/r) where A is a constant ⇒r′=v=(√(A−((2C)/r))) ⇒(dr/( (√(A−((2C)/r)))))=dt ⇒∫(dr/( (√(A−((2C)/r)))))=∫dt=t+B, B=const. u=(√r)⇒r=u^2 and dr=2udu ∫(dr/( (√(A−((2C)/r)))))=∫((√r)/( (√(Ar−2C))))dr= =∫((u2udu)/( (√(Au^2 −2C))))=(1/( (√A)))∫u((2udu)/( (√(u^2 −((2C)/A)))))= =(1/( (√A)))∫u((d(u^2 −2C/A))/( (√(u^2 −2C/A))))= =(2/( (√A)))∫u d((√(u^2 −((2C)/A)))) IBP : (2/( (√A)))∫u d((√(u^2 −((2C)/A))))= =(2/( (√A)))(u(√(u^2 −((2C)/A)))−∫(√(u^2 −((2C)/A)))du)= =(√((8C)/A^2 ))(u(√(((√(A/(2C))) u)^2 −1))−∫(√(((√(A/(2C))) u)^2 −1))du) ∫(√(((√(A/(2C))) u)^2 −1))du= (w=(√(A/(2C)))u) =(√((2C)/A))∫(√(w^2 −1))dw w=secθ and dw=secθ tanθ dθ ∫(√(w^2 −1))dw=∫((sin^2 θ)/(cos^3 θ))dθ= =∫((sinθ)/(cos^3 θ))sinθdθ=(1/2)∫sinθd((1/(cos^2 θ))) IBP : ∫(√(w^2 −1))dw=(1/2)(((sinθ)/(cos^2 θ))−∫(dθ/(cosθ)))= =(1/2)(((sinθ)/(cos^2 θ))−ln(secθ+tanθ))= =(1/2)(tanθsecθ−ln(secθ+tanθ))= =(1/2)(w(√(w^2 −1))−ln(w+(√(w^2 −1)))) Now all integrals have been calculated! We need to just go back and plug every substitution so we get the final result ...
$${Observe}:\: \\ $$$${r}''=\frac{{d}}{{dt}}\left(\frac{{dr}}{{dt}}\right)=\frac{{dv}}{{dt}}=\frac{{dv}}{{dr}}\:\frac{{dr}}{{dt}}={v}\frac{{dv}}{{dr}}\:,\:{where}\:{v}=\frac{{dr}}{{dt}} \\ $$$${Then}: \\ $$$${vdv}=\frac{{Cdr}}{{r}^{\mathrm{2}} }=−{Cd}\left(\frac{\mathrm{1}}{{r}}\right)=−{d}\left(\frac{{C}}{{r}}\right) \\ $$$${vdv}=\frac{\mathrm{1}}{\mathrm{2}}{d}\left({v}^{\mathrm{2}} \right)={d}\left(\frac{{v}^{\mathrm{2}} }{\mathrm{2}}\right) \\ $$$$\Rightarrow{d}\left(\frac{{v}^{\mathrm{2}} }{\mathrm{2}}\right)=−{d}\left(\frac{{C}}{{r}}\right) \\ $$$${d}\left(\frac{{v}^{\mathrm{2}} }{\mathrm{2}}\:+\:\frac{{C}}{{r}}\right)=\mathrm{0}\Rightarrow\frac{{v}^{\mathrm{2}} }{\mathrm{2}}=\frac{{A}}{\mathrm{2}}−\frac{{C}}{{r}} \\ $$$${where}\:{A}\:{is}\:{a}\:{constant} \\ $$$$\Rightarrow{r}'={v}=\sqrt{{A}−\frac{\mathrm{2}{C}}{{r}}} \\ $$$$\Rightarrow\frac{{dr}}{\:\sqrt{{A}−\frac{\mathrm{2}{C}}{{r}}}}={dt} \\ $$$$\Rightarrow\int\frac{{dr}}{\:\sqrt{{A}−\frac{\mathrm{2}{C}}{{r}}}}=\int{dt}={t}+{B},\:{B}={const}. \\ $$$${u}=\sqrt{{r}}\Rightarrow{r}={u}^{\mathrm{2}} \:{and}\:{dr}=\mathrm{2}{udu} \\ $$$$\int\frac{{dr}}{\:\sqrt{{A}−\frac{\mathrm{2}{C}}{{r}}}}=\int\frac{\sqrt{{r}}}{\:\sqrt{{Ar}−\mathrm{2}{C}}}{dr}= \\ $$$$=\int\frac{{u}\mathrm{2}{udu}}{\:\sqrt{{Au}^{\mathrm{2}} −\mathrm{2}{C}}}=\frac{\mathrm{1}}{\:\sqrt{{A}}}\int{u}\frac{\mathrm{2}{udu}}{\:\sqrt{{u}^{\mathrm{2}} −\frac{\mathrm{2}{C}}{{A}}}}= \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{A}}}\int{u}\frac{{d}\left({u}^{\mathrm{2}} −\mathrm{2}{C}/{A}\right)}{\:\sqrt{{u}^{\mathrm{2}} −\mathrm{2}{C}/{A}}}= \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{{A}}}\int{u}\:{d}\left(\sqrt{{u}^{\mathrm{2}} −\frac{\mathrm{2}{C}}{{A}}}\right) \\ $$$${IBP}\:: \\ $$$$\frac{\mathrm{2}}{\:\sqrt{{A}}}\int{u}\:{d}\left(\sqrt{{u}^{\mathrm{2}} −\frac{\mathrm{2}{C}}{{A}}}\right)= \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{{A}}}\left({u}\sqrt{{u}^{\mathrm{2}} −\frac{\mathrm{2}{C}}{{A}}}−\int\sqrt{{u}^{\mathrm{2}} −\frac{\mathrm{2}{C}}{{A}}}{du}\right)= \\ $$$$=\sqrt{\frac{\mathrm{8}{C}}{{A}^{\mathrm{2}} }}\left({u}\sqrt{\left(\sqrt{\frac{{A}}{\mathrm{2}{C}}}\:{u}\right)^{\mathrm{2}} −\mathrm{1}}−\int\sqrt{\left(\sqrt{\frac{{A}}{\mathrm{2}{C}}}\:{u}\right)^{\mathrm{2}} −\mathrm{1}}{du}\right) \\ $$$$\int\sqrt{\left(\sqrt{\frac{{A}}{\mathrm{2}{C}}}\:{u}\right)^{\mathrm{2}} −\mathrm{1}}{du}=\:\left({w}=\sqrt{\frac{{A}}{\mathrm{2}{C}}}{u}\right) \\ $$$$=\sqrt{\frac{\mathrm{2}{C}}{{A}}}\int\sqrt{{w}^{\mathrm{2}} −\mathrm{1}}{dw} \\ $$$${w}={sec}\theta\:{and}\:{dw}={sec}\theta\:{tan}\theta\:{d}\theta \\ $$$$\int\sqrt{{w}^{\mathrm{2}} −\mathrm{1}}{dw}=\int\frac{{sin}^{\mathrm{2}} \theta}{{cos}^{\mathrm{3}} \theta}{d}\theta= \\ $$$$=\int\frac{{sin}\theta}{{cos}^{\mathrm{3}} \theta}{sin}\theta{d}\theta=\frac{\mathrm{1}}{\mathrm{2}}\int{sin}\theta{d}\left(\frac{\mathrm{1}}{{cos}^{\mathrm{2}} \theta}\right) \\ $$$${IBP}\:: \\ $$$$\int\sqrt{{w}^{\mathrm{2}} −\mathrm{1}}{dw}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{sin}\theta}{{cos}^{\mathrm{2}} \theta}−\int\frac{{d}\theta}{{cos}\theta}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{sin}\theta}{{cos}^{\mathrm{2}} \theta}−{ln}\left({sec}\theta+{tan}\theta\right)\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({tan}\theta{sec}\theta−{ln}\left({sec}\theta+{tan}\theta\right)\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left({w}\sqrt{{w}^{\mathrm{2}} −\mathrm{1}}−{ln}\left({w}+\sqrt{{w}^{\mathrm{2}} −\mathrm{1}}\right)\right) \\ $$$${Now}\:{all}\:{integrals}\:{have}\:{been}\:{calculated}! \\ $$$${We}\:{need}\:{to}\:{just}\:{go}\:{back}\:{and}\:{plug} \\ $$$${every}\:{substitution}\:{so}\:{we}\:{get}\:{the}\:{final} \\ $$$${result} \\ $$$$… \\ $$

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