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Question Number 81554 by jagoll last updated on 14/Feb/20

y′′′ +4y′ = 3x−1 what is solution

$${y}'''\:+\mathrm{4}{y}'\:=\:\mathrm{3}{x}−\mathrm{1} \\ $$$${what}\:{is}\:{solution} \\ $$

Answered by TANMAY PANACEA last updated on 14/Feb/20

let y=e^(mx) be a solution so we get y^′ =(dy/dx)=me^(mx) →y^(′′) =m^2 e^(mx) y^(′′′) =m^3 e^(mx) complemenary function m^3 e^(mx) +4me^(mx) =0 e^(mx) ×m(m^2 +4)=0 e^(mx) ≠0 so eithethe m=0 or m=2i or −2i C.F=Ae^(0.x) +Be^(2ix) +Ce^(−2ix) (A,B and C constant) now calculation of particular intregal y=((3x−1)/(D^3 +4D)) where D=(d/dx) y=(1/D).(1/(D^2 +4)).(3x−1) y=(4+D^2 )^(−1) ×(((3x^2 )/2)−x) y={4(1+(D^2 /4))}^(−1) ×(((3x^2 )/2)−x) =(1/4)(1−(D^2 /4)+(D^4 /(16))+...)(((3x^2 )/2)−x) note (1+p)^(−1) =1−p+p^2 −p^3 +...∞ calculation (1/4)(((3x^2 )/2)−x)−(D^2 /(16))(((3x^2 )/2)−x)+(D^4 /(64))(((3x^2 )/2)−x)..othester is zero ((3x^2 )/8)−(x/4)−(3/(32))×2 =((3x^2 )/8)−(x/4)−(3/(16)) so compete solution y=Ae^(0.x) +Be^(2ix) +Ce^(−2ix) +((3x^2 )/8)−(x/4)−(3/(16)) note..(1)we can put e^(ix) =cosx+isinx e^(−ix) =cosx−isinx and (A−(3/4))=K new constant pls check

$${let}\:{y}={e}^{{mx}} \:\:{be}\:{a}\:{solution}\:{so}\:{we}\:{get} \\ $$$${y}^{'} =\frac{{dy}}{{dx}}={me}^{{mx}} \rightarrow{y}^{''} ={m}^{\mathrm{2}} {e}^{{mx}} \\ $$$${y}^{'''} ={m}^{\mathrm{3}} {e}^{{mx}} \\ $$$${complemenary}\:{function} \\ $$$${m}^{\mathrm{3}} {e}^{{mx}} +\mathrm{4}{me}^{{mx}} =\mathrm{0} \\ $$$${e}^{{mx}} ×{m}\left({m}^{\mathrm{2}} +\mathrm{4}\right)=\mathrm{0} \\ $$$${e}^{{mx}} \neq\mathrm{0}\:\:{so} \\ $$$${eithethe}\:{m}=\mathrm{0}\:{or}\:{m}=\mathrm{2}{i}\:{or}\:−\mathrm{2}{i} \\ $$$$\boldsymbol{{C}}.\boldsymbol{{F}}={Ae}^{\mathrm{0}.{x}} +\boldsymbol{{B}}{e}^{\mathrm{2}{ix}} +\boldsymbol{{C}}{e}^{−\mathrm{2}{ix}} \:\:\left({A},{B}\:{and}\:{C}\:{constant}\right) \\ $$$${now} \\ $$$${calculation}\:{of}\:{particular}\:{intregal} \\ $$$${y}=\frac{\mathrm{3}{x}−\mathrm{1}}{{D}^{\mathrm{3}} +\mathrm{4}{D}}\:\:\:{where}\:{D}=\frac{{d}}{{dx}} \\ $$$${y}=\frac{\mathrm{1}}{{D}}.\frac{\mathrm{1}}{{D}^{\mathrm{2}} +\mathrm{4}}.\left(\mathrm{3}{x}−\mathrm{1}\right) \\ $$$${y}=\left(\mathrm{4}+{D}^{\mathrm{2}} \right)^{−\mathrm{1}} ×\left(\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}−{x}\right) \\ $$$${y}=\left\{\mathrm{4}\left(\mathrm{1}+\frac{{D}^{\mathrm{2}} }{\mathrm{4}}\right)\right\}^{−\mathrm{1}} ×\left(\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}−{x}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}−\frac{{D}^{\mathrm{2}} }{\mathrm{4}}+\frac{{D}^{\mathrm{4}} }{\mathrm{16}}+...\right)\left(\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}−{x}\right) \\ $$$${note} \\ $$$$\left(\mathrm{1}+{p}\right)^{−\mathrm{1}} =\mathrm{1}−{p}+{p}^{\mathrm{2}} −{p}^{\mathrm{3}} +...\infty \\ $$$$ \\ $$$${calculation} \\ $$$$\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}−{x}\right)−\frac{{D}^{\mathrm{2}} }{\mathrm{16}}\left(\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}−{x}\right)+\frac{{D}^{\mathrm{4}} }{\mathrm{64}}\left(\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{2}}−{x}\right)..{othester}\:{is}\:{zero} \\ $$$$\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{8}}−\frac{{x}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{32}}×\mathrm{2} \\ $$$$=\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{8}}−\frac{{x}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{16}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$${so}\:{compete}\:{solution} \\ $$$${y}={Ae}^{\mathrm{0}.{x}} +{Be}^{\mathrm{2}{ix}} +{Ce}^{−\mathrm{2}{ix}} +\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{8}}−\frac{{x}}{\mathrm{4}}−\frac{\mathrm{3}}{\mathrm{16}} \\ $$$$\boldsymbol{{note}}..\left(\mathrm{1}\right)\boldsymbol{{we}}\:\boldsymbol{{can}}\:\boldsymbol{{put}}\:\boldsymbol{{e}}^{\boldsymbol{{ix}}} =\boldsymbol{{cosx}}+\boldsymbol{{isinx}} \\ $$$$\boldsymbol{{e}}^{−\boldsymbol{{ix}}} =\boldsymbol{{cosx}}−\boldsymbol{{isinx}} \\ $$$$\boldsymbol{{and}}\:\left(\boldsymbol{{A}}−\frac{\mathrm{3}}{\mathrm{4}}\right)={K}\:\:{new}\:{constant} \\ $$$${pls}\:{check} \\ $$$$ \\ $$$$ \\ $$

Commented by jagoll last updated on 14/Feb/20

can we get grades A,B and C mister?

$${can}\:{we}\:{get}\:{grades}\:{A},{B}\:{and}\:{C}\:{mister}? \\ $$

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(0)=... y′(1)=... y′′(2)=...

$${you}\:{gave}\:{only}\:{a}\:{d}.{e}.\:{without}\:{initial} \\ $$$${conditions},\:{so}\:{you}\:{can}\:{not}\:{determine} \\ $$$${these}\:{constants}.\:{to}\:{determine}\:{A},{B},{C} \\ $$$${you}\:{need}\:{three}\:{initial}\:{conditions}\:{such}\:{as} \\ $$$${y}\left(\mathrm{0}\right)=... \\ $$$${y}'\left(\mathrm{1}\right)=... \\ $$$${y}''\left(\mathrm{2}\right)=... \\ $$

Commented by jagoll last updated on 14/Feb/20

oo yes sir thank you

$${oo}\:{yes}\:{sir}\:{thank}\:{you} \\ $$

Answered by Joel578 last updated on 14/Feb/20

• homogeneous solution with characteristic eq. λ^3 + 4λ = 0 ⇒ λ_1 = 0, λ_(2,3) = ± 2i ⇒ y_h (x) = C_1 + C_2 cos 2x + C_3 sin 2x • particular solution y_p = Ax^2 + Bx + C y_p ′ = 2Ax + B y_p ′′ = 2A y_p ′′′ = 0 Substitute to original question 0 + 4(2Ax + B) = 3x − 1 ⇒ { ((8A = 3)),((4B = −1)) :} ⇒ { ((A = (3/8))),((B = ((−1)/4))) :} ⇒ y_p ′ = (3/4)x − (1/4) ⇒ y_p = (3/8)x^2 − (1/4)x + K ∴ y(x) = y_h (x) + y_p (x) = C + C_2 cos 2x + C_3 sin 2x + (3/8)x^2 − (1/4)x

$$\bullet\:\mathrm{homogeneous}\:\mathrm{solution}\:\mathrm{with}\:\mathrm{characteristic}\:\mathrm{eq}. \\ $$$$\:\:\:\:\lambda^{\mathrm{3}} \:+\:\mathrm{4}\lambda\:=\:\mathrm{0}\:\Rightarrow\:\lambda_{\mathrm{1}} \:=\:\mathrm{0},\:\lambda_{\mathrm{2},\mathrm{3}} \:=\:\pm\:\mathrm{2}{i} \\ $$$$\:\:\:\:\Rightarrow\:{y}_{{h}} \left({x}\right)\:=\:{C}_{\mathrm{1}} \:+\:{C}_{\mathrm{2}} \:\mathrm{cos}\:\mathrm{2}{x}\:+\:{C}_{\mathrm{3}} \:\mathrm{sin}\:\mathrm{2}{x} \\ $$$$ \\ $$$$\bullet\:\mathrm{particular}\:\mathrm{solution} \\ $$$$\:\:\:\:{y}_{{p}} \:=\:{Ax}^{\mathrm{2}} \:+\:{Bx}\:+\:{C} \\ $$$$\:\:\:{y}_{{p}} '\:=\:\mathrm{2}{Ax}\:+\:{B} \\ $$$$\:\:{y}_{{p}} ''\:=\:\mathrm{2}{A} \\ $$$$\:{y}_{{p}} '''\:=\:\mathrm{0} \\ $$$$\:\:\mathrm{Substitute}\:\mathrm{to}\:\mathrm{original}\:\mathrm{question} \\ $$$$\:\:\mathrm{0}\:+\:\mathrm{4}\left(\mathrm{2}{Ax}\:+\:{B}\right)\:=\:\mathrm{3}{x}\:−\:\mathrm{1} \\ $$$$\Rightarrow\:\begin{cases}{\mathrm{8}{A}\:=\:\mathrm{3}}\\{\mathrm{4}{B}\:=\:−\mathrm{1}}\end{cases}\:\Rightarrow\:\begin{cases}{{A}\:=\:\frac{\mathrm{3}}{\mathrm{8}}}\\{{B}\:=\:\frac{−\mathrm{1}}{\mathrm{4}}}\end{cases} \\ $$$$\Rightarrow\:{y}_{{p}} '\:=\:\frac{\mathrm{3}}{\mathrm{4}}{x}\:−\:\frac{\mathrm{1}}{\mathrm{4}}\:\:\Rightarrow\:{y}_{{p}} \:=\:\frac{\mathrm{3}}{\mathrm{8}}{x}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{4}}{x}\:+\:{K} \\ $$$$ \\ $$$$\therefore\:{y}\left({x}\right)\:=\:{y}_{{h}} \left({x}\right)\:+\:{y}_{{p}} \left({x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{C}\:+\:{C}_{\mathrm{2}} \:\mathrm{cos}\:\mathrm{2}{x}\:+\:{C}_{\mathrm{3}} \:\mathrm{sin}\:\mathrm{2}{x}\:+\:\frac{\mathrm{3}}{\mathrm{8}}{x}^{\mathrm{2}} \:−\:\frac{\mathrm{1}}{\mathrm{4}}{x} \\ $$

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