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Question Number 197792 by Mastermind last updated on 28/Sep/23
Solve the following differential equation 1) y′′ + y = e^x + x^3 , y(0)=2, y′(0)=0 2) y′′ + y^′ − 2y = x + sin2x, y(0)=1, y′(0)=0 3) y′′ − y′ = xe^x , y(0)=2, y′(0)= 1 Thank you
$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{following}\:\mathrm{differential}\:\mathrm{equation} \\ $$$$\left.\mathrm{1}\right)\:\mathrm{y}''\:+\:\mathrm{y}\:=\:\mathrm{e}^{\mathrm{x}} \:+\:\mathrm{x}^{\mathrm{3}} ,\:\:\:\:\:\:\:\:\:\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{2},\:\mathrm{y}'\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:\mathrm{y}''\:+\:\mathrm{y}^{'} \:−\:\mathrm{2y}\:=\:\mathrm{x}\:+\:\mathrm{sin2x},\:\:\:\:\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{1},\:\mathrm{y}'\left(\mathrm{0}\right)=\mathrm{0} \\ $$$$\left.\mathrm{3}\right)\:\mathrm{y}''\:−\:\mathrm{y}'\:=\:\mathrm{xe}^{\mathrm{x}} ,\:\:\:\:\:\:\:\:\:\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{2},\:\mathrm{y}'\left(\mathrm{0}\right)=\:\mathrm{1} \\ $$$$ \\ $$$$ \\ $$$$\mathrm{Thank}\:\mathrm{you} \\ $$
Answered by qaz last updated on 29/Sep/23
1)(y′−y)′+(y′−y)=e^x +x^3 2)(y′−y)′+2(y′−y)=x+sin 2x 3)y′′−y′=xe^x −−−−−−− y′+p(x)y=q(x) ⇒y=e^(−∫p(x)dx) (C+∫q(x)e^(∫p(x)dx) dx)
$$\left.\mathrm{1}\right)\left({y}'−{y}\right)'+\left({y}'−{y}\right)={e}^{{x}} +{x}^{\mathrm{3}} \\ $$$$\left.\mathrm{2}\right)\left({y}'−{y}\right)'+\mathrm{2}\left({y}'−{y}\right)={x}+\mathrm{sin}\:\mathrm{2}{x} \\ $$$$\left.\mathrm{3}\right){y}''−{y}'={xe}^{{x}} \\ $$$$−−−−−−− \\ $$$${y}'+{p}\left({x}\right){y}={q}\left({x}\right)\:\Rightarrow{y}={e}^{−\int{p}\left({x}\right){dx}} \left({C}+\int{q}\left({x}\right){e}^{\int{p}\left({x}\right){dx}} {dx}\right) \\ $$
Answered by Tokugami last updated on 30/Sep/23
Use Laplace Transform: 1. y′′+y=e^x +x^3 s^2 Y(s)−sy(0)−y′(0)+Y(s)=(1/(s−1))+(6/s^4 ) (s^2 +1)Y(s)−2s−0=(1/(s−1))+(6/s^4 ) Y(s)=(1/((s−1)(s^2 +1)))+(6/(s^4 (s^2 +1)))+((2s)/(s^2 +1)) Y(s)=((1/2)/(s−1))+((−(1/2)s−(1/2))/(s^2 +1))−(6/s^2 )+(6/s^4 )+(6/(s^2 +1))+2((s/(s^2 +1))) L^(−1) {Y(s)}=L^(−1) {(1/2)((1/(s−1)))+(3/2)((s/(s^2 +1)))+((11)/2)((1/(s^2 +1)))−(6/s^2 )+(6/s^4 )} y=(1/2)e^x +(3/2)cos(x)+((11)/2)sin(x)−6x+(1/(20))x^5 2. y′′+y′−2y=x+sin(2x), y(0)=1, y′(0)=0 s^2 Y(s)−sy(0)−y′(0)+sY(s)−y(0)−2Y(s)=(1/s^2 )+(2/(s^2 +4)) (s^2 +s−2)Y(s)−s−1=(1/s^2 )+(2/(s^2 +4)) (s^2 +s−2)Y(s)=(1/s^2 )+(2/(s^2 +4))+s+1 Y(s)=((s^5 +s^4 +4s^3 +7s^2 +4)/(s^2 (s^2 +4)(s−1)(s+2)))=((−1/2)/s^2 )−((1/4)/s)+((1/6)/(s+2))+((17/15)/(s−1))−(1/(20))(((s+6)/(s^2 +4))) L^(−1) {Y(s)}=L^(−1) {−(1/4)((1/s))+((17)/(15))((1/(s−1)))+(1/6)((1/(s+2)))−(1/2)((1/s^2 ))−(1/(20))((s/(s^2 +4)))−(3/(20))((2/(s^2 +4)))} y=−(1/4)−(1/2)x+((17)/(15))e^x +(1/6)e^(−2x) −(1/(20))cos(2x)−(3/(20))sin(2x) 3. y′′−y′=xe^x , y(0)=2, y′(0)=1 s^2 Y(s)−sy(0)−y′(0)−sY(s)+y(0)=(1/((s−1)^2 )) (s^2 −s)Y(s)−2s+1=(1/((s−1)^2 )) Y(s)=(1/(s(s−1)^3 ))+((2s−1)/(s(s−1))) Y(s)=((s^2 −3s+3)/((s−1)^3 ))+(1/(s−1))=(2/(s−1))−(1/((s−1)^2 ))+(1/((s−1)^3 )) Inverse Laplace: y=(2−x+(1/2)x^2 )e^x
$$\mathrm{Use}\:\mathrm{Laplace}\:\mathrm{Transform}: \\ $$$$\mathrm{1}.\:{y}''+{y}={e}^{{x}} +{x}^{\mathrm{3}} \\ $$$${s}^{\mathrm{2}} {Y}\left({s}\right)−{sy}\left(\mathrm{0}\right)−{y}'\left(\mathrm{0}\right)+{Y}\left({s}\right)=\frac{\mathrm{1}}{{s}−\mathrm{1}}+\frac{\mathrm{6}}{{s}^{\mathrm{4}} } \\ $$$$\left({s}^{\mathrm{2}} +\mathrm{1}\right){Y}\left({s}\right)−\mathrm{2}{s}−\mathrm{0}=\frac{\mathrm{1}}{{s}−\mathrm{1}}+\frac{\mathrm{6}}{{s}^{\mathrm{4}} } \\ $$$${Y}\left({s}\right)=\frac{\mathrm{1}}{\left({s}−\mathrm{1}\right)\left({s}^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{6}}{{s}^{\mathrm{4}} \left({s}^{\mathrm{2}} +\mathrm{1}\right)}+\frac{\mathrm{2}{s}}{{s}^{\mathrm{2}} +\mathrm{1}} \\ $$$${Y}\left({s}\right)=\frac{\mathrm{1}/\mathrm{2}}{{s}−\mathrm{1}}+\frac{−\frac{\mathrm{1}}{\mathrm{2}}{s}−\frac{\mathrm{1}}{\mathrm{2}}}{{s}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{6}}{{s}^{\mathrm{2}} }+\frac{\mathrm{6}}{{s}^{\mathrm{4}} }+\frac{\mathrm{6}}{{s}^{\mathrm{2}} +\mathrm{1}}+\mathrm{2}\left(\frac{{s}}{{s}^{\mathrm{2}} +\mathrm{1}}\right) \\ $$$$\mathscr{L}^{−\mathrm{1}} \left\{{Y}\left({s}\right)\right\}=\mathscr{L}^{−\mathrm{1}} \left\{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{s}−\mathrm{1}}\right)+\frac{\mathrm{3}}{\mathrm{2}}\left(\frac{{s}}{{s}^{\mathrm{2}} +\mathrm{1}}\right)+\frac{\mathrm{11}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{s}^{\mathrm{2}} +\mathrm{1}}\right)−\frac{\mathrm{6}}{{s}^{\mathrm{2}} }+\frac{\mathrm{6}}{{s}^{\mathrm{4}} }\right\} \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{2}}{e}^{{x}} +\frac{\mathrm{3}}{\mathrm{2}}\mathrm{cos}\left({x}\right)+\frac{\mathrm{11}}{\mathrm{2}}\mathrm{sin}\left({x}\right)−\mathrm{6}{x}+\frac{\mathrm{1}}{\mathrm{20}}{x}^{\mathrm{5}} \\ $$$$\mathrm{2}.\:{y}''+{y}'−\mathrm{2}{y}={x}+\mathrm{sin}\left(\mathrm{2}{x}\right),\:{y}\left(\mathrm{0}\right)=\mathrm{1},\:{y}'\left(\mathrm{0}\right)=\mathrm{0} \\ $$$${s}^{\mathrm{2}} {Y}\left({s}\right)−{sy}\left(\mathrm{0}\right)−{y}'\left(\mathrm{0}\right)+{sY}\left({s}\right)−{y}\left(\mathrm{0}\right)−\mathrm{2}{Y}\left({s}\right)=\frac{\mathrm{1}}{{s}^{\mathrm{2}} }+\frac{\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{4}} \\ $$$$\left({s}^{\mathrm{2}} +{s}−\mathrm{2}\right){Y}\left({s}\right)−{s}−\mathrm{1}=\frac{\mathrm{1}}{{s}^{\mathrm{2}} }+\frac{\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{4}} \\ $$$$\left({s}^{\mathrm{2}} +{s}−\mathrm{2}\right){Y}\left({s}\right)=\frac{\mathrm{1}}{{s}^{\mathrm{2}} }+\frac{\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{4}}+{s}+\mathrm{1} \\ $$$${Y}\left({s}\right)=\frac{{s}^{\mathrm{5}} +{s}^{\mathrm{4}} +\mathrm{4}{s}^{\mathrm{3}} +\mathrm{7}{s}^{\mathrm{2}} +\mathrm{4}}{{s}^{\mathrm{2}} \left({s}^{\mathrm{2}} +\mathrm{4}\right)\left({s}−\mathrm{1}\right)\left({s}+\mathrm{2}\right)}=\frac{−\mathrm{1}/\mathrm{2}}{{s}^{\mathrm{2}} }−\frac{\mathrm{1}/\mathrm{4}}{{s}}+\frac{\mathrm{1}/\mathrm{6}}{{s}+\mathrm{2}}+\frac{\mathrm{17}/\mathrm{15}}{{s}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{20}}\left(\frac{{s}+\mathrm{6}}{{s}^{\mathrm{2}} +\mathrm{4}}\right) \\ $$$$\mathscr{L}^{−\mathrm{1}} \left\{{Y}\left({s}\right)\right\}=\mathscr{L}^{−\mathrm{1}} \left\{−\frac{\mathrm{1}}{\mathrm{4}}\left(\frac{\mathrm{1}}{{s}}\right)+\frac{\mathrm{17}}{\mathrm{15}}\left(\frac{\mathrm{1}}{{s}−\mathrm{1}}\right)+\frac{\mathrm{1}}{\mathrm{6}}\left(\frac{\mathrm{1}}{{s}+\mathrm{2}}\right)−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{s}^{\mathrm{2}} }\right)−\frac{\mathrm{1}}{\mathrm{20}}\left(\frac{{s}}{{s}^{\mathrm{2}} +\mathrm{4}}\right)−\frac{\mathrm{3}}{\mathrm{20}}\left(\frac{\mathrm{2}}{{s}^{\mathrm{2}} +\mathrm{4}}\right)\right\} \\ $$$${y}=−\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}{x}+\frac{\mathrm{17}}{\mathrm{15}}{e}^{{x}} +\frac{\mathrm{1}}{\mathrm{6}}{e}^{−\mathrm{2}{x}} −\frac{\mathrm{1}}{\mathrm{20}}\mathrm{cos}\left(\mathrm{2}{x}\right)−\frac{\mathrm{3}}{\mathrm{20}}\mathrm{sin}\left(\mathrm{2}{x}\right) \\ $$$$\mathrm{3}.\:{y}''−{y}'={xe}^{{x}} ,\:{y}\left(\mathrm{0}\right)=\mathrm{2},\:{y}'\left(\mathrm{0}\right)=\mathrm{1} \\ $$$${s}^{\mathrm{2}} {Y}\left({s}\right)−{sy}\left(\mathrm{0}\right)−{y}'\left(\mathrm{0}\right)−{sY}\left({s}\right)+{y}\left(\mathrm{0}\right)=\frac{\mathrm{1}}{\left({s}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\left({s}^{\mathrm{2}} −{s}\right){Y}\left({s}\right)−\mathrm{2}{s}+\mathrm{1}=\frac{\mathrm{1}}{\left({s}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${Y}\left({s}\right)=\frac{\mathrm{1}}{{s}\left({s}−\mathrm{1}\right)^{\mathrm{3}} }+\frac{\mathrm{2}{s}−\mathrm{1}}{{s}\left({s}−\mathrm{1}\right)} \\ $$$${Y}\left({s}\right)=\frac{{s}^{\mathrm{2}} −\mathrm{3}{s}+\mathrm{3}}{\left({s}−\mathrm{1}\right)^{\mathrm{3}} }+\frac{\mathrm{1}}{{s}−\mathrm{1}}=\frac{\mathrm{2}}{{s}−\mathrm{1}}−\frac{\mathrm{1}}{\left({s}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\left({s}−\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\mathrm{Inverse}\:\mathrm{Laplace}: \\ $$$${y}=\left(\mathrm{2}−{x}+\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} \right){e}^{{x}} \\ $$$$ \\ $$
Commented by AST last updated on 03/Oct/23
I guess he′s talking about the “boss” aspect.
$${I}\:{guess}\:{he}'{s}\:{talking}\:{about}\:{the}\:“{boss}''\:{aspect}. \\ $$
Commented by Mastermind last updated on 30/Sep/23
Thank you so much my BOSS
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{my}\:\mathrm{BOSS} \\ $$
Commented by necx122 last updated on 02/Oct/23
only Nigerians speak this way
$${only}\:{Nigerians}\:{speak}\:{this}\:{way} \\ $$
Commented by Mastermind last updated on 02/Oct/23
No, not only Nigerians... English is an official language
$$\mathrm{No},\:\mathrm{not}\:\mathrm{only}\:\mathrm{Nigerians}…\:\mathrm{English}\:\mathrm{is}\:\mathrm{an}\:\mathrm{official}\:\mathrm{language} \\ $$
Commented by Mastermind last updated on 04/Oct/23
Oh! Ok
$$\mathrm{Oh}!\:\mathrm{Ok} \\ $$

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