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Question Number 81586 by jagoll last updated on 14/Feb/20

Dear mister Tanmay y ′′ − y′ −6y = 2sin 3x find solution let y = e^(mx) ? or not

$${Dear}\:{mister}\: \\ $$$${Tanmay} \\ $$$${y}\:''\:−\:{y}'\:−\mathrm{6}{y}\:=\:\mathrm{2sin}\:\mathrm{3}{x}\: \\ $$$${find}\:{solution} \\ $$$$\: \\ $$$${let}\:{y}\:=\:{e}^{{mx}} \:?\:{or}\:{not}\: \\ $$$$ \\ $$

Commented by jagoll last updated on 14/Feb/20

y′ =me^(mx) ⇒y′′ = m^2 e^(mx) m^2 e^(mx) −me^x −6e^(mx) = 0 e^(mx) (m^2 −m−6)=0 (m−3)(m+2)=0 ⇒m=3 ∧ m=−2 y = Ae^(3x) + Be^(−2x)

$${y}'\:={me}^{{mx}} \:\Rightarrow{y}''\:=\:{m}^{\mathrm{2}} \:{e}^{{mx}} \\ $$$${m}^{\mathrm{2}} {e}^{{mx}} −{me}^{{x}} −\mathrm{6}{e}^{{mx}} \:=\:\mathrm{0} \\ $$$${e}^{{mx}} \:\left({m}^{\mathrm{2}} −{m}−\mathrm{6}\right)=\mathrm{0} \\ $$$$\left({m}−\mathrm{3}\right)\left({m}+\mathrm{2}\right)=\mathrm{0}\:\Rightarrow{m}=\mathrm{3}\:\wedge\:{m}=−\mathrm{2} \\ $$$${y}\:=\:{Ae}^{\mathrm{3}{x}} \:+\:{Be}^{−\mathrm{2}{x}} \: \\ $$$$ \\ $$

Commented by jagoll last updated on 14/Feb/20

particular integral y = ((2sin 3x)/(D^2 −D−6))

$${particular}\:{integral}\: \\ $$$${y}\:=\:\frac{\mathrm{2sin}\:\mathrm{3}{x}}{{D}^{\mathrm{2}} −{D}−\mathrm{6}} \\ $$

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

y=2.((sin3x)/(D^2 −6−D)) =2.(((D^2 −6+D)sin3x)/((D^2 −6)^2 −D^2 )) =2.(((−9sin3x−6sin3x+3cos3x))/((−3^2 −6)^2 −(−3^2 ))) =(2/(225+9))×((3cos3x−15sin3x)/1) =(2/(234))×(3cos3x−15sin3x)

$${y}=\mathrm{2}.\frac{{sin}\mathrm{3}{x}}{{D}^{\mathrm{2}} −\mathrm{6}−{D}} \\ $$$$=\mathrm{2}.\frac{\left({D}^{\mathrm{2}} −\mathrm{6}+{D}\right){sin}\mathrm{3}{x}}{\left({D}^{\mathrm{2}} −\mathrm{6}\right)^{\mathrm{2}} −{D}^{\mathrm{2}} } \\ $$$$=\mathrm{2}.\frac{\left(−\mathrm{9}{sin}\mathrm{3}{x}−\mathrm{6}{sin}\mathrm{3}{x}+\mathrm{3}{cos}\mathrm{3}{x}\right)}{\left(−\mathrm{3}^{\mathrm{2}} −\mathrm{6}\right)^{\mathrm{2}} −\left(−\mathrm{3}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{2}}{\mathrm{225}+\mathrm{9}}×\frac{\mathrm{3}{cos}\mathrm{3}{x}−\mathrm{15}{sin}\mathrm{3}{x}}{\mathrm{1}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{234}}×\left(\mathrm{3}{cos}\mathrm{3}{x}−\mathrm{15}{sin}\mathrm{3}{x}\right) \\ $$

Commented by jagoll last updated on 14/Feb/20

sir how to get line 2?

$${sir}\:{how}\:{to}\:{get}\:{line}\:\mathrm{2}? \\ $$

Commented by jagoll last updated on 14/Feb/20

oo by multiply conjugate (D^2 −6)+D

$${oo}\:{by}\:{multiply}\:\:{conjugate} \\ $$$$\left({D}^{\mathrm{2}} −\mathrm{6}\right)+{D}\: \\ $$

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

yes ...objective to multiply by conjuage to get denominator in terms of D^2

$${yes}\:...{objective}\:{to}\:{multiply}\:{by}\:{conjuage}\:{to}\:{get} \\ $$$${denominator}\:{in}\:{terms}\:{of}\:{D}^{\mathrm{2}} \\ $$$$ \\ $$

Answered by Joel578 last updated on 14/Feb/20

• homogeneous solution with characteristic eq. λ^2 − λ − 6 = 0 ⇒ λ_1 = −2, λ_2 = 3 ⇒ y_h (x) = C_1 e^(−2x) + C_2 e^(3x) • particular solution y_p (x) = A cos 3x + B sin 3x y_p ′(x) = −3A sin 3x + 3B cos 3x y_p ′′(x) = −9A cos 3x − 9B sin 3x Substitute to the question (−15A − 3B)cos 3x + (−15B + 3A)sin 3x = 2 sin 3x ⇒ { ((−15A − 3B = 0)),((−15B + 3A = 2)) :} ⇒ { ((A = (1/(39)))),((B = −(5/(39)))) :} ⇒ y_p = (1/(39)) cos 3x − (5/(39)) sin 3x ∴ y(x) = y_h (x) + y_p (x) = C_1 e^(−2x) + C_2 e^(3x) + (1/(39)) cos 3x − (5/(39)) sin 3x

$$\bullet\:\mathrm{homogeneous}\:\mathrm{solution}\:\mathrm{with}\:\mathrm{characteristic}\:\mathrm{eq}. \\ $$$$\:\:\:\lambda^{\mathrm{2}} \:−\:\lambda\:−\:\mathrm{6}\:=\:\mathrm{0}\:\Rightarrow\:\lambda_{\mathrm{1}} \:=\:−\mathrm{2},\:\lambda_{\mathrm{2}} \:=\:\mathrm{3} \\ $$$$\:\:\Rightarrow\:{y}_{{h}} \left({x}\right)\:=\:{C}_{\mathrm{1}} {e}^{−\mathrm{2}{x}} \:+\:{C}_{\mathrm{2}} {e}^{\mathrm{3}{x}} \\ $$$$ \\ $$$$\bullet\:\mathrm{particular}\:\mathrm{solution} \\ $$$${y}_{{p}} \left({x}\right)\:=\:{A}\:\mathrm{cos}\:\mathrm{3}{x}\:+\:{B}\:\mathrm{sin}\:\mathrm{3}{x} \\ $$$${y}_{{p}} '\left({x}\right)\:=\:−\mathrm{3}{A}\:\mathrm{sin}\:\mathrm{3}{x}\:+\:\mathrm{3}{B}\:\mathrm{cos}\:\mathrm{3}{x}\:\: \\ $$$${y}_{{p}} ''\left({x}\right)\:=\:−\mathrm{9}{A}\:\mathrm{cos}\:\mathrm{3}{x}\:−\:\mathrm{9}{B}\:\mathrm{sin}\:\mathrm{3}{x} \\ $$$$\mathrm{Substitute}\:\mathrm{to}\:\mathrm{the}\:\mathrm{question} \\ $$$$\left(−\mathrm{15}{A}\:−\:\mathrm{3}{B}\right)\mathrm{cos}\:\mathrm{3}{x}\:+\:\left(−\mathrm{15}{B}\:+\:\mathrm{3}{A}\right)\mathrm{sin}\:\mathrm{3}{x}\:=\:\mathrm{2}\:\mathrm{sin}\:\mathrm{3}{x} \\ $$$$\Rightarrow\:\begin{cases}{−\mathrm{15}{A}\:−\:\mathrm{3}{B}\:=\:\mathrm{0}}\\{−\mathrm{15}{B}\:+\:\mathrm{3}{A}\:=\:\mathrm{2}}\end{cases}\:\Rightarrow\:\begin{cases}{{A}\:=\:\frac{\mathrm{1}}{\mathrm{39}}}\\{{B}\:=\:−\frac{\mathrm{5}}{\mathrm{39}}}\end{cases} \\ $$$$\Rightarrow\:{y}_{{p}} \:=\:\frac{\mathrm{1}}{\mathrm{39}}\:\mathrm{cos}\:\mathrm{3}{x}\:−\:\frac{\mathrm{5}}{\mathrm{39}}\:\mathrm{sin}\:\mathrm{3}{x} \\ $$$$ \\ $$$$\therefore\:{y}\left({x}\right)\:=\:{y}_{{h}} \left({x}\right)\:+\:{y}_{{p}} \left({x}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:{C}_{\mathrm{1}} \:{e}^{−\mathrm{2}{x}} \:+\:{C}_{\mathrm{2}} \:{e}^{\mathrm{3}{x}} \:+\:\frac{\mathrm{1}}{\mathrm{39}}\:\mathrm{cos}\:\mathrm{3}{x}\:−\:\frac{\mathrm{5}}{\mathrm{39}}\:\mathrm{sin}\:\mathrm{3}{x} \\ $$

Commented by jagoll last updated on 14/Feb/20

thank you mister

$${thank}\:{you}\:{mister} \\ $$

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