Question Number 81554 by jagoll last updated on 14/Feb/20
$${y}'''\:+\mathrm{4}{y}'\:=\:\mathrm{3}{x}−\mathrm{1} \\ $$$${what}\:{is}\:{solution} \\ $$
Answered by TANMAY PANACEA last updated on 14/Feb/20
$${let}\:{y}={e}^{{mx}} \:\:{be}\:{a}\:{solution}\:{so}\:{we}\:{get} \\ $$$${y}^{'} =\frac{{dy}}{{dx}}={me}^{{mx}} \rightarrow{y}^{''} ={m}^{\mathrm{2}} {e}^{{mx}} \\ $$$${y}^{'''} ={m}^{\mathrm{3}} {e}^{{mx}} \\ $$$${complemenary}\:{function} \\ $$$${m}^{\mathrm{3}} {e}^{{mx}} +\mathrm{4}{me}^{{mx}} =\mathrm{0} \\ $$$${e}^{{mx}} ×{m}\left({m}^{\mathrm{2}} +\mathrm{4}\right)=\mathrm{0} \\ $$$${e}^{{mx}} \neq\mathrm{0}\:\:{so} \\ $$$${eithethe}\:{m}=\mathrm{0}\:{or}\:{m}=\mathrm{2}{i}\:{or}\:−\mathrm{2}{i} \\ $$$$\boldsymbol{{C}}.\boldsymbol{{F}}={Ae}^{\mathrm{0}.{x}} +\boldsymbol{{B}}{e}^{\mathrm{2}{ix}} +\boldsymbol{{C}}{e}^{−\mathrm{2}{ix}} \:\:\left({A},{B}\:{and}\:{C}\:{constant}\right) \\ $$$${now} \\ $$$${calculation}\:{of}\:{particular}\:{intregal} \\ $$$${y}=\frac{\mathrm{3}{x}−\mathrm{1}}{{D}^{\mathrm{3}} +\mathrm{4}{D}}\:\:\:{where}\:{D}=\frac{{d}}{{dx}} \\ $$$${y}=\frac{\mathrm{1}}{{D}}.\frac{\mathrm{1}}{{D}^{\mathrm{2}} +\mathrm{4}}.\left(\mathrm{3}{x}−\mathrm{1}\right) \\ $$$${y}=\left(\mathrm{4}+{D}^{\mathrm{2}} \right)^{−\mathrm{1}} ×\left(\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}−{x}\right) \\ $$$${y}=\left\{\mathrm{4}\left(\mathrm{1}+\frac{{D}^{\mathrm{2}} }{\mathrm{4}}\right)\right\}^{−\mathrm{1}} ×\left(\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}−{x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}−\frac{{D}^{\mathrm{2}} }{\mathrm{4}}+\frac{{D}^{\mathrm{4}} }{\mathrm{16}}+…\right)\left(\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}−{x}\right) \\ $$$${note} \\ $$$$\left(\mathrm{1}+{p}\right)^{−\mathrm{1}} =\mathrm{1}−{p}+{p}^{\mathrm{2}} −{p}^{\mathrm{3}} +…\infty \\ $$$$ \\ $$$${calculation} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}−{x}\right)−\frac{{D}^{\mathrm{2}} }{\mathrm{16}}\left(\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}−{x}\right)+\frac{{D}^{\mathrm{4}} }{\mathrm{64}}\left(\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}−{x}\right)..{othester}\:{is}\:{zero} \\ $$$$\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{8}}−\frac{{x}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{32}}×\mathrm{2} \\ $$$$=\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{8}}−\frac{{x}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{16}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${so}\:{compete}\:{solution} \\ $$$${y}={Ae}^{\mathrm{0}.{x}} +{Be}^{\mathrm{2}{ix}} +{Ce}^{−\mathrm{2}{ix}} +\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{8}}−\frac{{x}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{16}} \\ $$$$\boldsymbol{{note}}..\left(\mathrm{1}\right)\boldsymbol{{we}}\:\boldsymbol{{can}}\:\boldsymbol{{put}}\:\boldsymbol{{e}}^{\boldsymbol{{ix}}} =\boldsymbol{{cosx}}+\boldsymbol{{isinx}} \\ $$$$\boldsymbol{{e}}^{−\boldsymbol{{ix}}} =\boldsymbol{{cosx}}−\boldsymbol{{isinx}} \\ $$$$\boldsymbol{{and}}\:\left(\boldsymbol{{A}}−\frac{\mathrm{3}}{\mathrm{4}}\right)={K}\:\:{new}\:{constant} \\ $$$${pls}\:{check} \\ $$$$ \\ $$$$ \\ $$
Commented by jagoll last updated on 14/Feb/20
$${can}\:{we}\:{get}\:{grades}\:{A},{B}\:{and}\:{C}\:{mister}? \\ $$
Commented by mr W last updated on 14/Feb/20
$${you}\:{gave}\:{only}\:{a}\:{d}.{e}.\:{without}\:{initial} \\ $$$${conditions},\:{so}\:{you}\:{can}\:{not}\:{determine} \\ $$$${these}\:{constants}.\:{to}\:{determine}\:{A},{B},{C} \\ $$$${you}\:{need}\:{three}\:{initial}\:{conditions}\:{such}\:{as} \\ $$$${y}\left(\mathrm{0}\right)=… \\ $$$${y}'\left(\mathrm{1}\right)=… \\ $$$${y}''\left(\mathrm{2}\right)=… \\ $$
Commented by jagoll last updated on 14/Feb/20
$${oo}\:{yes}\:{sir}\:{thank}\:{you} \\ $$
Answered by Joel578 last updated on 14/Feb/20
$$\bullet\:\mathrm{homogeneous}\:\mathrm{solution}\:\mathrm{with}\:\mathrm{characteristic}\:\mathrm{eq}. \\ $$$$\:\:\:\:\lambda^{\mathrm{3}} \:+\:\mathrm{4}\lambda\:=\:\mathrm{0}\:\Rightarrow\:\lambda_{\mathrm{1}} \:=\:\mathrm{0},\:\lambda_{\mathrm{2},\mathrm{3}} \:=\:\pm\:\mathrm{2}{i} \\ $$$$\:\:\:\:\Rightarrow\:{y}_{{h}} \left({x}\right)\:=\:{C}_{\mathrm{1}} \:+\:{C}_{\mathrm{2}} \:\mathrm{cos}\:\mathrm{2}{x}\:+\:{C}_{\mathrm{3}} \:\mathrm{sin}\:\mathrm{2}{x} \\ $$$$ \\ $$$$\bullet\:\mathrm{particular}\:\mathrm{solution} \\ $$$$\:\:\:\:{y}_{{p}} \:=\:{Ax}^{\mathrm{2}} \:+\:{Bx}\:+\:{C} \\ $$$$\:\:\:{y}_{{p}} '\:=\:\mathrm{2}{Ax}\:+\:{B} \\ $$$$\:\:{y}_{{p}} ''\:=\:\mathrm{2}{A} \\ $$$$\:{y}_{{p}} '''\:=\:\mathrm{0} \\ $$$$\:\:\mathrm{Substitute}\:\mathrm{to}\:\mathrm{original}\:\mathrm{question} \\ $$$$\:\:\mathrm{0}\:+\:\mathrm{4}\left(\mathrm{2}{Ax}\:+\:{B}\right)\:=\:\mathrm{3}{x}\:−\:\mathrm{1} \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{8}{A}\:=\:\mathrm{3}}\\{\mathrm{4}{B}\:=\:−\mathrm{1}}\end{cases}\:\Rightarrow\:\begin{cases}{{A}\:=\:\frac{\mathrm{3}}{\mathrm{8}}}\\{{B}\:=\:\frac{−\mathrm{1}}{\mathrm{4}}}\end{cases} \\ $$$$\Rightarrow\:{y}_{{p}} '\:=\:\frac{\mathrm{3}}{\mathrm{4}}{x}\:−\:\frac{\mathrm{1}}{\mathrm{4}}\:\:\Rightarrow\:{y}_{{p}} \:=\:\frac{\mathrm{3}}{\mathrm{8}}{x}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{4}}{x}\:+\:{K} \\ $$$$ \\ $$$$\therefore\:{y}\left({x}\right)\:=\:{y}_{{h}} \left({x}\right)\:+\:{y}_{{p}} \left({x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{C}\:+\:{C}_{\mathrm{2}} \:\mathrm{cos}\:\mathrm{2}{x}\:+\:{C}_{\mathrm{3}} \:\mathrm{sin}\:\mathrm{2}{x}\:+\:\frac{\mathrm{3}}{\mathrm{8}}{x}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{4}}{x} \\ $$