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Question Number 55668 by mr W last updated on 01/Mar/19

Find all functions y=f(x) such that y′y′′=y′′′.

$${Find}\:{all}\:{functions}\:{y}={f}\left({x}\right)\:{such}\:{that} \\ $$$${y}'{y}''={y}'''. \\ $$

Commented by malwaan last updated on 02/Mar/19

$$\mathrm{y}=\mathrm{e}^{\mathrm{x}} \\ $$

Commented by mr W last updated on 02/Mar/19

please check sir: with y=e^x y′=e^x y′′=e^x y′′′=e^x y′y′′=e^x e^x =e^(2x) ≠y^(′′′) =e^x ⇒y=e^x is not a solution

$${please}\:{check}\:{sir}: \\ $$$${with}\:{y}={e}^{{x}} \\ $$$${y}'={e}^{{x}} \\ $$$${y}''={e}^{{x}} \\ $$$${y}'''={e}^{{x}} \\ $$$${y}'{y}''={e}^{{x}} {e}^{{x}} ={e}^{\mathrm{2}{x}} \neq{y}^{'''} ={e}^{{x}} \\ $$$$\Rightarrow{y}={e}^{{x}} \:{is}\:{not}\:{a}\:{solution} \\ $$

Commented by mr W last updated on 02/Mar/19

it′s obvious that y=c_1 x+c_2 is a solution, since y′=c_1 , y′′=0, y′′′=0⇒ y′y′′=y′′′=0. but are there any other solutions?

$${it}'{s}\:{obvious}\:{that}\:{y}={c}_{\mathrm{1}} {x}+{c}_{\mathrm{2}} \:{is}\:{a}\:{solution}, \\ $$$${since}\:{y}'={c}_{\mathrm{1}} ,\:{y}''=\mathrm{0},\:{y}'''=\mathrm{0}\Rightarrow\:{y}'{y}''={y}'''=\mathrm{0}. \\ $$$${but}\:{are}\:{there}\:{any}\:{other}\:{solutions}? \\ $$

Answered by mr W last updated on 03/Mar/19

let g(x)=y′=f′(x) ⇒gg′=g′′ let h(x)=g′(x) g′′=((dh(x))/dx)=((dh(x))/dg)×(dg/dx)=h(dh/dg) ⇒gh=h(dh/dg) if h≠0: (h=0⇒g(x)=c_1 ⇒y=c_1 x+c_2 ) ⇒gdg=dh ⇒∫gdg=∫dh ⇒((g^2 +c_1 )/2)=h=(dg/dx) ⇒(g^2 +c_1 )dx=2dg if g^2 +c_1 ≠0: ⇒(dg/(g^2 +c_1 ))=(dx/2) ⇒∫(dg/(g^2 +c_1 ))=∫(dx/2) if c_1 =0: ⇒∫(dg/g^2 )=∫(dx/2) ⇒−(1/g)=((x+c_2 )/2) ⇒g=y′=(dy/dx)=−(2/(x+c_2 )) ⇒dy=−((2dx)/(x+c_2 )) ⇒y=−2ln ∣x+c_2 ∣+c_3 if c_1 >0, say c_1 =c_4 ^2 ⇒∫(dg/(g^2 +c_4 ^2 ))=∫(dx/2) ⇒(1/c_4 )tan^(−1) (g/c_4 )=((x+c_5 )/2) ⇒g=y′=(dy/dx)=c_4 tan ((c_4 (x+c_5 ))/2) ⇒y=∫c_4 tan ((c_4 (x+c_5 ))/2)dx ⇒y=2∫tan ((c_4 (x+c_5 ))/2)d(((c_4 (x+c_5 ))/2)) ⇒y=−2ln ∣cos ((c_4 (x+c_5 ))/2)∣+c_6 if c_1 <0, say c_1 =−c_4 ^2 ⇒∫(dg/(g^2 −c_4 ^2 ))=∫(dx/2) ⇒(1/(2c_4 ))ln ∣((g−c_4 )/(g+c_4 ))∣=((x+c_5 )/2) ⇒ln ∣1−((2c_4 )/(g+c_4 ))∣=c_4 (x+c_5 ) ⇒1−((2c_4 )/(g+c_4 ))=e^(c_4 (x+c_5 )) ⇒((2c_4 )/(g+c_4 ))=1−e^(c_4 (x+c_5 )) ⇒g=y′=(dy/dx)=c_4 [(2/(1−e^(c_4 (x+c_5 )) ))−1] ⇒y=∫c_4 [(2/(1−e^(c_4 (x+c_5 )) ))−1]dx ⇒y=c_4 [∫((2dx)/(1−e^(c_4 (x+c_5 )) ))−x] ⇒y=c_4 [2(x+c_5 )−((2ln ∣1−e^(c_4 (x+c_5 )) ∣)/c_4 )−x] ⇒y=c_4 x−2ln ∣1−e^(c_4 (x+c_5 )) ∣+c_6 if g^2 +c_1 =0: if c_1 =0: ⇒g=y′=(dy/dx)=0 ⇒y=c_2 if c_1 >0: g^2 +c_1 ≠0 if c_1 <0: g^2 +c_1 =0 ⇒g^2 =−c_1 =c_2 ^2 say ⇒g=±c_2 ⇒g=(dy/dx)=±c_2 ⇒y=±c_2 x+c_3 summary: following functions are possible solutions: y=f(x)=c_1 x+c_2 y=f(x)=−2ln ∣x+c_1 ∣+c_2 y=f(x)=−2ln ∣cos (c_1 x+c_2 )∣+c_3 y=f(x)=−2ln ∣1−e^(c_1 x+c_2 ) ∣+c_1 x+c_3

$${let}\:{g}\left({x}\right)={y}'={f}'\left({x}\right) \\ $$$$\Rightarrow{gg}'={g}'' \\ $$$${let}\:{h}\left({x}\right)={g}'\left({x}\right) \\ $$$${g}''=\frac{{dh}\left({x}\right)}{{dx}}=\frac{{dh}\left({x}\right)}{{dg}}×\frac{{dg}}{{dx}}={h}\frac{{dh}}{{dg}} \\ $$$$\Rightarrow{gh}={h}\frac{{dh}}{{dg}} \\ $$$${if}\:{h}\neq\mathrm{0}:\:\left({h}=\mathrm{0}\Rightarrow{g}\left({x}\right)={c}_{\mathrm{1}} \Rightarrow{y}={c}_{\mathrm{1}} {x}+{c}_{\mathrm{2}} \right) \\ $$$$\Rightarrow{gdg}={dh} \\ $$$$\Rightarrow\int{gdg}=\int{dh} \\ $$$$\Rightarrow\frac{{g}^{\mathrm{2}} +{c}_{\mathrm{1}} }{\mathrm{2}}={h}=\frac{{dg}}{{dx}} \\ $$$$\Rightarrow\left({g}^{\mathrm{2}} +{c}_{\mathrm{1}} \right){dx}=\mathrm{2}{dg} \\ $$$$\boldsymbol{{if}}\:\boldsymbol{{g}}^{\mathrm{2}} +\boldsymbol{{c}}_{\mathrm{1}} \neq\mathrm{0}: \\ $$$$\Rightarrow\frac{{dg}}{{g}^{\mathrm{2}} +{c}_{\mathrm{1}} }=\frac{{dx}}{\mathrm{2}} \\ $$$$\Rightarrow\int\frac{{dg}}{{g}^{\mathrm{2}} +{c}_{\mathrm{1}} }=\int\frac{{dx}}{\mathrm{2}} \\ $$$${if}\:{c}_{\mathrm{1}} =\mathrm{0}: \\ $$$$\Rightarrow\int\frac{{dg}}{{g}^{\mathrm{2}} }=\int\frac{{dx}}{\mathrm{2}} \\ $$$$\Rightarrow−\frac{\mathrm{1}}{{g}}=\frac{{x}+{c}_{\mathrm{2}} }{\mathrm{2}} \\ $$$$\Rightarrow{g}={y}'=\frac{{dy}}{{dx}}=−\frac{\mathrm{2}}{{x}+{c}_{\mathrm{2}} } \\ $$$$\Rightarrow{dy}=−\frac{\mathrm{2}{dx}}{{x}+{c}_{\mathrm{2}} } \\ $$$$\Rightarrow{y}=−\mathrm{2ln}\:\mid{x}+{c}_{\mathrm{2}} \mid+{c}_{\mathrm{3}} \\ $$$${if}\:{c}_{\mathrm{1}} >\mathrm{0},\:{say}\:{c}_{\mathrm{1}} ={c}_{\mathrm{4}} ^{\mathrm{2}} \\ $$$$\Rightarrow\int\frac{{dg}}{{g}^{\mathrm{2}} +{c}_{\mathrm{4}} ^{\mathrm{2}} }=\int\frac{{dx}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{c}_{\mathrm{4}} }\mathrm{tan}^{−\mathrm{1}} \frac{{g}}{{c}_{\mathrm{4}} }=\frac{{x}+{c}_{\mathrm{5}} }{\mathrm{2}} \\ $$$$\Rightarrow{g}={y}'=\frac{{dy}}{{dx}}={c}_{\mathrm{4}} \mathrm{tan}\:\frac{{c}_{\mathrm{4}} \left({x}+{c}_{\mathrm{5}} \right)}{\mathrm{2}} \\ $$$$\Rightarrow{y}=\int{c}_{\mathrm{4}} \mathrm{tan}\:\frac{{c}_{\mathrm{4}} \left({x}+{c}_{\mathrm{5}} \right)}{\mathrm{2}}{dx} \\ $$$$\Rightarrow{y}=\mathrm{2}\int\mathrm{tan}\:\frac{{c}_{\mathrm{4}} \left({x}+{c}_{\mathrm{5}} \right)}{\mathrm{2}}{d}\left(\frac{{c}_{\mathrm{4}} \left({x}+{c}_{\mathrm{5}} \right)}{\mathrm{2}}\right) \\ $$$$\Rightarrow{y}=−\mathrm{2ln}\:\mid\mathrm{cos}\:\frac{{c}_{\mathrm{4}} \left({x}+{c}_{\mathrm{5}} \right)}{\mathrm{2}}\mid+{c}_{\mathrm{6}} \\ $$$${if}\:{c}_{\mathrm{1}} <\mathrm{0},\:{say}\:{c}_{\mathrm{1}} =−{c}_{\mathrm{4}} ^{\mathrm{2}} \\ $$$$\Rightarrow\int\frac{{dg}}{{g}^{\mathrm{2}} −{c}_{\mathrm{4}} ^{\mathrm{2}} }=\int\frac{{dx}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}{c}_{\mathrm{4}} }\mathrm{ln}\:\mid\frac{{g}−{c}_{\mathrm{4}} }{{g}+{c}_{\mathrm{4}} }\mid=\frac{{x}+{c}_{\mathrm{5}} }{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{ln}\:\mid\mathrm{1}−\frac{\mathrm{2}{c}_{\mathrm{4}} }{{g}+{c}_{\mathrm{4}} }\mid={c}_{\mathrm{4}} \left({x}+{c}_{\mathrm{5}} \right) \\ $$$$\Rightarrow\mathrm{1}−\frac{\mathrm{2}{c}_{\mathrm{4}} }{{g}+{c}_{\mathrm{4}} }={e}^{{c}_{\mathrm{4}} \left({x}+{c}_{\mathrm{5}} \right)} \\ $$$$\Rightarrow\frac{\mathrm{2}{c}_{\mathrm{4}} }{{g}+{c}_{\mathrm{4}} }=\mathrm{1}−{e}^{{c}_{\mathrm{4}} \left({x}+{c}_{\mathrm{5}} \right)} \\ $$$$\Rightarrow{g}={y}'=\frac{{dy}}{{dx}}={c}_{\mathrm{4}} \left[\frac{\mathrm{2}}{\mathrm{1}−{e}^{{c}_{\mathrm{4}} \left({x}+{c}_{\mathrm{5}} \right)} }−\mathrm{1}\right] \\ $$$$\Rightarrow{y}=\int{c}_{\mathrm{4}} \left[\frac{\mathrm{2}}{\mathrm{1}−{e}^{{c}_{\mathrm{4}} \left({x}+{c}_{\mathrm{5}} \right)} }−\mathrm{1}\right]{dx} \\ $$$$\Rightarrow{y}={c}_{\mathrm{4}} \left[\int\frac{\mathrm{2}{dx}}{\mathrm{1}−{e}^{{c}_{\mathrm{4}} \left({x}+{c}_{\mathrm{5}} \right)} }−{x}\right] \\ $$$$\Rightarrow{y}={c}_{\mathrm{4}} \left[\mathrm{2}\left({x}+{c}_{\mathrm{5}} \right)−\frac{\mathrm{2ln}\:\mid\mathrm{1}−{e}^{{c}_{\mathrm{4}} \left({x}+{c}_{\mathrm{5}} \right)} \mid}{{c}_{\mathrm{4}} }−{x}\right] \\ $$$$\Rightarrow{y}={c}_{\mathrm{4}} {x}−\mathrm{2ln}\:\mid\mathrm{1}−{e}^{{c}_{\mathrm{4}} \left({x}+{c}_{\mathrm{5}} \right)} \mid+{c}_{\mathrm{6}} \\ $$$$\boldsymbol{{if}}\:\boldsymbol{{g}}^{\mathrm{2}} +\boldsymbol{{c}}_{\mathrm{1}} =\mathrm{0}: \\ $$$${if}\:{c}_{\mathrm{1}} =\mathrm{0}: \\ $$$$\Rightarrow{g}={y}'=\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\Rightarrow{y}={c}_{\mathrm{2}} \\ $$$${if}\:{c}_{\mathrm{1}} >\mathrm{0}:\:{g}^{\mathrm{2}} +{c}_{\mathrm{1}} \neq\mathrm{0} \\ $$$${if}\:{c}_{\mathrm{1}} <\mathrm{0}:\:{g}^{\mathrm{2}} +{c}_{\mathrm{1}} =\mathrm{0} \\ $$$$\Rightarrow{g}^{\mathrm{2}} =−{c}_{\mathrm{1}} ={c}_{\mathrm{2}} ^{\mathrm{2}} \:{say} \\ $$$$\Rightarrow{g}=\pm{c}_{\mathrm{2}} \\ $$$$\Rightarrow{g}=\frac{{dy}}{{dx}}=\pm{c}_{\mathrm{2}} \\ $$$$\Rightarrow{y}=\pm{c}_{\mathrm{2}} {x}+{c}_{\mathrm{3}} \\ $$$$ \\ $$$${summary}: \\ $$$${following}\:{functions}\:{are}\:{possible}\:{solutions}: \\ $$$${y}={f}\left({x}\right)={c}_{\mathrm{1}} {x}+{c}_{\mathrm{2}} \\ $$$${y}={f}\left({x}\right)=−\mathrm{2ln}\:\mid{x}+{c}_{\mathrm{1}} \mid+{c}_{\mathrm{2}} \\ $$$${y}={f}\left({x}\right)=−\mathrm{2ln}\:\mid\mathrm{cos}\:\left({c}_{\mathrm{1}} {x}+{c}_{\mathrm{2}} \right)\mid+{c}_{\mathrm{3}} \\ $$$${y}={f}\left({x}\right)=−\mathrm{2ln}\:\mid\mathrm{1}−{e}^{{c}_{\mathrm{1}} {x}+{c}_{\mathrm{2}} } \mid+{c}_{\mathrm{1}} {x}+{c}_{\mathrm{3}} \\ $$

Commented by MJS last updated on 02/Mar/19