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Question Number 1023 by 123456 last updated on 20/May/15
φ_n :R→R n∈N^∗ φ_n =t^n (d^n φ/dt^n ) φ_1 (t)=?,φ_1 (1)=+1 φ_2 (t)=?,φ_2 (1)=−1,φ_2 ^′ (1)=+1 φ_3 (t)=?,φ_3 (1)=+1,φ_3 ^′ (1)=−1,φ_3 ^(′′) (1)=+1
$$\phi_{{n}} :\mathbb{R}\rightarrow\mathbb{R} \\ $$$${n}\in\mathbb{N}^{\ast} \\ $$$$\phi_{{n}} ={t}^{{n}} \frac{{d}^{{n}} \phi}{{dt}^{{n}} } \\ $$$$\phi_{\mathrm{1}} \left({t}\right)=?,\phi_{\mathrm{1}} \left(\mathrm{1}\right)=+\mathrm{1} \\ $$$$\phi_{\mathrm{2}} \left({t}\right)=?,\phi_{\mathrm{2}} \left(\mathrm{1}\right)=−\mathrm{1},\phi_{\mathrm{2}} ^{'} \left(\mathrm{1}\right)=+\mathrm{1} \\ $$$$\phi_{\mathrm{3}} \left({t}\right)=?,\phi_{\mathrm{3}} \left(\mathrm{1}\right)=+\mathrm{1},\phi_{\mathrm{3}} ^{'} \left(\mathrm{1}\right)=−\mathrm{1},\phi_{\mathrm{3}} ^{''} \left(\mathrm{1}\right)=+\mathrm{1} \\ $$
Commented by prakash jain last updated on 21/May/15
φ_1 (t)=t((dφ_1 (t))/dt)⇒ln y=ln t+C φ_1 (t)=kt φ_1 (1)=1⇒k=1 φ_1 (t)=t
$$\phi_{\mathrm{1}} \left({t}\right)={t}\frac{{d}\phi_{\mathrm{1}} \left({t}\right)}{{dt}}\Rightarrow\mathrm{ln}\:{y}=\mathrm{ln}\:{t}+{C} \\ $$$$\phi_{\mathrm{1}} \left({t}\right)={kt} \\ $$$$\phi_{\mathrm{1}} \left(\mathrm{1}\right)=\mathrm{1}\Rightarrow{k}=\mathrm{1} \\ $$$$\phi_{\mathrm{1}} \left({t}\right)={t} \\ $$
Answered by prakash jain last updated on 21/May/15
φ_2 (t)=t^2 (d^2 φ_2 /dt^2 ) t=e^x (dφ_2 /dt)=e^(−x) (dφ_2 /dx), (d^2 φ_2 /dt)=((d^2 φ_2 /dx^2 )−(dφ_2 /dx))e^(−2x) φ_2 =φ_2 ^(′′) −φ_2 ^′ φ_2 ^(′′) −φ_2 ^′ −φ_2 =0 characteristic equation r^2 −r−1=0 r_1 ,r_2 =((1±(√5))/2) φ_2 (x)=c_1 e^(r_1 x) +c_2 e^(r_2 x) φ_2 (t)=c_1 t^r_1 +c_2 t^r_2 φ_2 (t)=−1⇒c_1 +c_2 =−1 φ_2 ′=c_1 r_1 t^(r_1 −1) +c_2 r_2 t^(r_2 −1) φ_2 ′(1)=1 1=c_1 r_1 +c_2 r_2 =c_1 (((1+(√5))/2))−c_1 (((1−(√5))/2))−((1−(√5))/2) 1+((1−(√5))/2)=c_1 (√5)⇒c_1 =((3−(√5))/( (√5))) c_2 =−1−c_1 =−1−((3−(√5))/( (√5)))=((2(√5)−3)/2) φ_2 (t)=c_1 x^r_1 +c_2 x^r_2 , r_1 =((1+(√5))/2), r_2 =((1−(√5))/2)
$$\phi_{\mathrm{2}} \left({t}\right)={t}^{\mathrm{2}} \frac{{d}^{\mathrm{2}} \phi_{\mathrm{2}} }{{dt}^{\mathrm{2}} } \\ $$$${t}={e}^{{x}} \\ $$$$\frac{{d}\phi_{\mathrm{2}} }{{dt}}={e}^{−{x}} \frac{{d}\phi_{\mathrm{2}} }{{dx}},\:\frac{{d}^{\mathrm{2}} \phi_{\mathrm{2}} }{{dt}}=\left(\frac{{d}^{\mathrm{2}} \phi_{\mathrm{2}} }{{dx}^{\mathrm{2}} }−\frac{{d}\phi_{\mathrm{2}} }{{dx}}\right){e}^{−\mathrm{2}{x}} \\ $$$$\phi_{\mathrm{2}} =\phi_{\mathrm{2}} ^{''} −\phi_{\mathrm{2}} ^{'} \\ $$$$\phi_{\mathrm{2}} ^{''} −\phi_{\mathrm{2}} ^{'} −\phi_{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{characteristic}\:\mathrm{equation} \\ $$$${r}^{\mathrm{2}} −{r}−\mathrm{1}=\mathrm{0} \\ $$$${r}_{\mathrm{1}} ,{r}_{\mathrm{2}} =\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\phi_{\mathrm{2}} \left({x}\right)={c}_{\mathrm{1}} {e}^{{r}_{\mathrm{1}} {x}} +{c}_{\mathrm{2}} {e}^{{r}_{\mathrm{2}} {x}} \\ $$$$\phi_{\mathrm{2}} \left({t}\right)={c}_{\mathrm{1}} {t}^{{r}_{\mathrm{1}} } +{c}_{\mathrm{2}} {t}^{{r}_{\mathrm{2}} } \\ $$$$\phi_{\mathrm{2}} \left({t}\right)=−\mathrm{1}\Rightarrow{c}_{\mathrm{1}} +{c}_{\mathrm{2}} =−\mathrm{1} \\ $$$$\phi_{\mathrm{2}} '={c}_{\mathrm{1}} {r}_{\mathrm{1}} {t}^{{r}_{\mathrm{1}} −\mathrm{1}} +{c}_{\mathrm{2}} {r}_{\mathrm{2}} {t}^{{r}_{\mathrm{2}} −\mathrm{1}} \\ $$$$\phi_{\mathrm{2}} '\left(\mathrm{1}\right)=\mathrm{1} \\ $$$$\mathrm{1}={c}_{\mathrm{1}} {r}_{\mathrm{1}} +{c}_{\mathrm{2}} {r}_{\mathrm{2}} ={c}_{\mathrm{1}} \left(\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}}\right)−{c}_{\mathrm{1}} \left(\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}\right)−\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\mathrm{1}+\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}}={c}_{\mathrm{1}} \sqrt{\mathrm{5}}\Rightarrow{c}_{\mathrm{1}} =\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{5}}} \\ $$$${c}_{\mathrm{2}} =−\mathrm{1}−{c}_{\mathrm{1}} =−\mathrm{1}−\frac{\mathrm{3}−\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{5}}}=\frac{\mathrm{2}\sqrt{\mathrm{5}}−\mathrm{3}}{\mathrm{2}} \\ $$$$\phi_{\mathrm{2}} \left({t}\right)={c}_{\mathrm{1}} {x}^{{r}_{\mathrm{1}} } +{c}_{\mathrm{2}} {x}^{{r}_{\mathrm{2}} } ,\:{r}_{\mathrm{1}} =\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}},\:{r}_{\mathrm{2}} =\frac{\mathrm{1}−\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

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