Question Number 49358 by peter frank last updated on 06/Dec/18
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18
![1)9(d^2 x/dt^2 )+6(dx/dt)+x=−50sint step1 9(d^2 x/dt^2 )+6(dx/dt)+x=0 x=e^(αt) is a solutiin 9(α^2 e^(αt) )+6(αe^(αt) )+e^(αt) =0 e^(αt) ≠0 9α^2 +6α+1=0 (3α+1)^2 =0 α=((−1)/3) C.F =Ate^((−t)/3) +Be^((−t)/3) =e^((−t)/3) (At+B) P.I calculation... D=(d/dt) D^2 =(d^2 /dt^2 ) (9D^2 +6D+1)x=−50sint x=((−50)/(9D^2 +6D+1))×sint =−50×((9D^2 +1−6D)/((9D^2 +1)^2 −36D^2 ))×sint D(sint)=cost D^2 (sint)=−sint =−50×((−9sint+sint−6cost)/((9×−1^2 +1)^2 −36(−1^2 ))) =−50×((−8sint−6cost)/(64+36)) =(1/2)×(8sint+6cost)=4sint+3cost 5((4/(5 ))sint+(3/5)cost) [ (4/5)=sinα (3/5)=cosα] =5cos(t−α) complete solution is x=e^((−t)/3) (At+B)+5cos(t−α)](https://www.tinkutara.com/question/Q49465.png)
$$\left.\mathrm{1}\right)\mathrm{9}\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }+\mathrm{6}\frac{{dx}}{{dt}}+{x}=−\mathrm{50}{sint} \\ $$$${step}\mathrm{1} \\ $$$$\mathrm{9}\frac{{d}^{\mathrm{2}} {x}}{{dt}^{\mathrm{2}} }+\mathrm{6}\frac{{dx}}{{dt}}+{x}=\mathrm{0} \\ $$$${x}={e}^{\alpha{t}} \:{is}\:{a}\:{solutiin} \\ $$$$\mathrm{9}\left(\alpha^{\mathrm{2}} {e}^{\alpha{t}} \right)+\mathrm{6}\left(\alpha{e}^{\alpha{t}} \right)+{e}^{\alpha{t}} =\mathrm{0} \\ $$$${e}^{\alpha{t}} \neq\mathrm{0} \\ $$$$\mathrm{9}\alpha^{\mathrm{2}} +\mathrm{6}\alpha+\mathrm{1}=\mathrm{0} \\ $$$$\left(\mathrm{3}\alpha+\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\alpha=\frac{−\mathrm{1}}{\mathrm{3}} \\ $$$${C}.{F}\:={Ate}^{\frac{−{t}}{\mathrm{3}}} +{Be}^{\frac{−{t}}{\mathrm{3}}} ={e}^{\frac{−{t}}{\mathrm{3}}} \left({At}+{B}\right) \\ $$$${P}.{I}\:{calculation}… \\ $$$${D}=\frac{{d}}{{dt}}\:\:\:\:\:{D}^{\mathrm{2}} =\frac{{d}^{\mathrm{2}} }{{dt}^{\mathrm{2}} } \\ $$$$\left(\mathrm{9}{D}^{\mathrm{2}} +\mathrm{6}{D}+\mathrm{1}\right){x}=−\mathrm{50}{sint} \\ $$$${x}=\frac{−\mathrm{50}}{\mathrm{9}{D}^{\mathrm{2}} +\mathrm{6}{D}+\mathrm{1}}×{sint} \\ $$$$=−\mathrm{50}×\frac{\mathrm{9}{D}^{\mathrm{2}} +\mathrm{1}−\mathrm{6}{D}}{\left(\mathrm{9}{D}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{36}{D}^{\mathrm{2}} }×{sint} \\ $$$${D}\left({sint}\right)={cost}\:\:\:{D}^{\mathrm{2}} \left({sint}\right)=−{sint} \\ $$$$ \\ $$$$=−\mathrm{50}×\frac{−\mathrm{9}{sint}+{sint}−\mathrm{6}{cost}}{\left(\mathrm{9}×−\mathrm{1}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} −\mathrm{36}\left(−\mathrm{1}^{\mathrm{2}} \right)} \\ $$$$=−\mathrm{50}×\frac{−\mathrm{8}{sint}−\mathrm{6}{cost}}{\mathrm{64}+\mathrm{36}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×\left(\mathrm{8}{sint}+\mathrm{6}{cost}\right)=\mathrm{4}{sint}+\mathrm{3}{cost} \\ $$$$\mathrm{5}\left(\frac{\mathrm{4}}{\mathrm{5}\:}{sint}+\frac{\mathrm{3}}{\mathrm{5}}{cost}\right)\:\left[\:\frac{\mathrm{4}}{\mathrm{5}}={sin}\alpha\:\:\:\frac{\mathrm{3}}{\mathrm{5}}={cos}\alpha\right] \\ $$$$=\mathrm{5}{cos}\left({t}−\alpha\right) \\ $$$${complete}\:{solution}\:{is} \\ $$$${x}={e}^{\frac{−{t}}{\mathrm{3}}} \left({At}+{B}\right)+\mathrm{5}{cos}\left({t}−\alpha\right) \\ $$$$ \\ $$
Commented by peter frank last updated on 07/Dec/18
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$$$ \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 07/Dec/18
$${most}\:{welcome}… \\ $$