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Question Number 2160 by Yozzis last updated on 05/Nov/15
Solve the d.e sin((d^3 y/dt^3 ))+3t^2 y=6t.
$${Solve}\:{the}\:{d}.{e} \\ $$$${sin}\left(\frac{{d}^{\mathrm{3}} {y}}{{dt}^{\mathrm{3}} }\right)+\mathrm{3}{t}^{\mathrm{2}} {y}=\mathrm{6}{t}. \\ $$
Commented by Yozzi last updated on 07/Nov/15
I was helping a friend with his engineering math homework and he therefore took a picture of the problem. It was a worksheet. I answered the question but but in that picture I saw a number of strange D.Es as part of another exercise. The exercise was to simply identify the type of D.E but I thought, ′How would one solve it or prove it is unsolvable?′ The D.E problem above this problem was also from the set of strange D.Es I saw. Yozzis=Yozzi
$${I}\:{was}\:{helping}\:{a}\:{friend}\:{with}\:{his}\: \\ $$$${engineering}\:{math}\:{homework}\:{and}\: \\ $$$${he}\:{therefore}\:{took}\:{a}\:{picture}\:{of}\:{the}\:{problem}. \\ $$$${It}\:{was}\:{a}\:{worksheet}.\:{I}\:{answered}\:{the}\:{question}\:{but} \\ $$$${but}\:{in}\:{that}\:{picture}\:{I}\:{saw}\:{a}\:{number}\:{of}\: \\ $$$${strange}\:{D}.{Es}\:{as}\:{part}\:{of}\:{another}\:{exercise}.\:{The}\:{exercise}\:{was}\:{to}\:{simply} \\ $$$${identify}\:{the}\:{type}\:{of}\:{D}.{E}\:{but}\:{I}\:{thought}, \\ $$$$'{How}\:{would}\:{one}\:{solve}\:{it}\:{or}\:{prove}\:{it}\:{is} \\ $$$${unsolvable}?'\:{The}\:{D}.{E}\:{problem}\:{above} \\ $$$${this}\:{problem}\:{was}\:{also}\:{from}\:{the}\:{set} \\ $$$${of}\:{strange}\:{D}.{Es}\:{I}\:{saw}. \\ $$$${Yozzis}={Yozzi} \\ $$
Commented by prakash jain last updated on 07/Nov/15
Difficult. Did you come across this while working on physics problem?
$$\mathrm{Difficult}.\:\mathrm{Did}\:\mathrm{you}\:\mathrm{come}\:\mathrm{across}\:\mathrm{this}\:\mathrm{while} \\ $$$$\mathrm{working}\:\mathrm{on}\:\mathrm{physics}\:\mathrm{problem}? \\ $$
Commented by prakash jain last updated on 08/Nov/15
assuming y can be differentiated 4 times cos ((d^3 y/dt^4 ))(d^4 y/dt^4 )=6−6ty−3t^2 (dy/dt) cos ((d^3 y/dt^4 ))=((d^4 y/dt^4 ))^(−1) (6−6ty−3t^2 (dy/dt)) 1=36t^2 +((d^4 y/dt^4 ))^((−2)) (6−6ty−3t^2 (dy/dt))^2 The above equation appears solvable with series expansion.
$${assuming}\:{y}\:{can}\:{be}\:{differentiated}\:\mathrm{4}\:{times} \\ $$$$\mathrm{cos}\:\left(\frac{{d}^{\mathrm{3}} {y}}{{dt}^{\mathrm{4}} }\right)\frac{{d}^{\mathrm{4}} {y}}{{dt}^{\mathrm{4}} }=\mathrm{6}−\mathrm{6}{ty}−\mathrm{3}{t}^{\mathrm{2}} \:\frac{{dy}}{{dt}} \\ $$$$\mathrm{cos}\:\left(\frac{{d}^{\mathrm{3}} {y}}{{dt}^{\mathrm{4}} }\right)=\left(\frac{{d}^{\mathrm{4}} {y}}{{dt}^{\mathrm{4}} }\right)^{−\mathrm{1}} \left(\mathrm{6}−\mathrm{6}{ty}−\mathrm{3}{t}^{\mathrm{2}} \:\frac{{dy}}{{dt}}\right) \\ $$$$\mathrm{1}=\mathrm{36}{t}^{\mathrm{2}} +\left(\frac{{d}^{\mathrm{4}} {y}}{{dt}^{\mathrm{4}} }\right)^{\left(−\mathrm{2}\right)} \left(\mathrm{6}−\mathrm{6}{ty}−\mathrm{3}{t}^{\mathrm{2}} \frac{{dy}}{{dt}}\right)^{\mathrm{2}} \\ $$$$\mathrm{The}\:\mathrm{above}\:\mathrm{equation}\:\mathrm{appears}\:\mathrm{solvable}\:\mathrm{with} \\ $$$$\mathrm{series}\:\mathrm{expansion}. \\ $$
Answered by prakash jain last updated on 08/Nov/15
Trying with variable substitution to remove sin function of a differnetial. y=(2/t)−(1/(3t^2 ))sin u sin ((d^3 y/dt^3 ))=6t−3t^2 y=6t−6t+sin u (d^3 y/dt^3 )= u computing (d^3 y/dt^3 ) in terms of u and t (dy/dt)=−(2/t^2 )+(2/(3t^3 ))sin u−(1/(3t^2 ))cos u(du/dt) (d^2 y/dt^2 )=(4/t^3 )+(2/(3t^3 ))cos u(du/dt)−(2/t^4 )sin u+(2/(3t^3 ))cos u(du/dt) −(1/(3t^2 ))cos u(d^2 u/dt^2 )+(1/(3t^2 ))sin u((du/dt))^2 =(4/t^3 )+(4/(3t^3 ))cos u(du/dt)−(2/t^4 )sin u−(1/(3t^2 ))cos u(d^2 u/dt^2 )+(1/(3t^2 ))sin u((du/dt))^2 (d^3 y/dt^3 )=−((12)/t^4 )−(4/t^4 )cos u(du/dt)+(4/(3t^3 ))cos u(d^2 u/dt^2 )−(4/(3t^3 ))sin u((du/dt))^2 +(8/t^5 )sin u−(2/t^4 )cos u(du/dt) +(2/(3t^3 ))cos u(d^2 u/dt^3 )+(1/(3t^2 ))sin u((du/dt))−(1/(3t^2 ))cos u(d^3 u/dt^3 ) +(4/t^4 )sin u((du/dt))^2 −(4/(3t^3 ))cos u((du/dt))^3 −(4/(3t^3 ))sin u(2(du/dt))((d^2 u/dt^2 )) we get an equation between t and u. still a very difficult equation
$$\mathrm{Trying}\:\mathrm{with}\:\mathrm{variable}\:\mathrm{substitution} \\ $$$$\mathrm{to}\:\mathrm{remove}\:\mathrm{sin}\:\mathrm{function}\:\mathrm{of}\:\mathrm{a}\:\mathrm{differnetial}. \\ $$$${y}=\frac{\mathrm{2}}{{t}}−\frac{\mathrm{1}}{\mathrm{3}{t}^{\mathrm{2}} }\mathrm{sin}\:{u} \\ $$$$\mathrm{sin}\:\left(\frac{{d}^{\mathrm{3}} {y}}{{dt}^{\mathrm{3}} }\right)=\mathrm{6}{t}−\mathrm{3}{t}^{\mathrm{2}} {y}=\mathrm{6}{t}−\mathrm{6}{t}+\mathrm{sin}\:{u} \\ $$$$\frac{{d}^{\mathrm{3}} {y}}{{dt}^{\mathrm{3}} }=\:{u} \\ $$$${computing}\:\frac{{d}^{\mathrm{3}} {y}}{{dt}^{\mathrm{3}} }\:{in}\:{terms}\:{of}\:{u}\:{and}\:{t} \\ $$$$\frac{{dy}}{{dt}}=−\frac{\mathrm{2}}{{t}^{\mathrm{2}} }+\frac{\mathrm{2}}{\mathrm{3}{t}^{\mathrm{3}} }\mathrm{sin}\:{u}−\frac{\mathrm{1}}{\mathrm{3}{t}^{\mathrm{2}} }\mathrm{cos}\:{u}\frac{{du}}{{dt}} \\ $$$$\frac{{d}^{\mathrm{2}} {y}}{{dt}^{\mathrm{2}} }=\frac{\mathrm{4}}{{t}^{\mathrm{3}} }+\frac{\mathrm{2}}{\mathrm{3}{t}^{\mathrm{3}} }\mathrm{cos}\:{u}\frac{{du}}{{dt}}−\frac{\mathrm{2}}{{t}^{\mathrm{4}} }\mathrm{sin}\:{u}+\frac{\mathrm{2}}{\mathrm{3}{t}^{\mathrm{3}} }\mathrm{cos}\:{u}\frac{{du}}{{dt}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{3}{t}^{\mathrm{2}} }\mathrm{cos}\:{u}\frac{{d}^{\mathrm{2}} {u}}{{dt}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}{t}^{\mathrm{2}} }\mathrm{sin}\:{u}\left(\frac{{du}}{{dt}}\right)^{\mathrm{2}} \\ $$$$=\frac{\mathrm{4}}{{t}^{\mathrm{3}} }+\frac{\mathrm{4}}{\mathrm{3}{t}^{\mathrm{3}} }\mathrm{cos}\:{u}\frac{{du}}{{dt}}−\frac{\mathrm{2}}{{t}^{\mathrm{4}} }\mathrm{sin}\:{u}−\frac{\mathrm{1}}{\mathrm{3}{t}^{\mathrm{2}} }\mathrm{cos}\:{u}\frac{{d}^{\mathrm{2}} {u}}{{dt}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3}{t}^{\mathrm{2}} }\mathrm{sin}\:{u}\left(\frac{{du}}{{dt}}\right)^{\mathrm{2}} \\ $$$$\frac{{d}^{\mathrm{3}} {y}}{{dt}^{\mathrm{3}} }=−\frac{\mathrm{12}}{{t}^{\mathrm{4}} }−\frac{\mathrm{4}}{{t}^{\mathrm{4}} }\mathrm{cos}\:{u}\frac{{du}}{{dt}}+\frac{\mathrm{4}}{\mathrm{3}{t}^{\mathrm{3}} }\mathrm{cos}\:{u}\frac{{d}^{\mathrm{2}} {u}}{{dt}^{\mathrm{2}} }−\frac{\mathrm{4}}{\mathrm{3}{t}^{\mathrm{3}} }\mathrm{sin}\:{u}\left(\frac{{du}}{{dt}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:+\frac{\mathrm{8}}{{t}^{\mathrm{5}} }\mathrm{sin}\:{u}−\frac{\mathrm{2}}{{t}^{\mathrm{4}} }\mathrm{cos}\:{u}\frac{{du}}{{dt}} \\ $$$$\:\:\:\:\:\:\:\:\:+\frac{\mathrm{2}}{\mathrm{3}{t}^{\mathrm{3}} }\mathrm{cos}\:{u}\frac{{d}^{\mathrm{2}} {u}}{{dt}^{\mathrm{3}} }+\frac{\mathrm{1}}{\mathrm{3}{t}^{\mathrm{2}} }\mathrm{sin}\:{u}\left(\frac{{du}}{{dt}}\right)−\frac{\mathrm{1}}{\mathrm{3}{t}^{\mathrm{2}} }\mathrm{cos}\:{u}\frac{{d}^{\mathrm{3}} {u}}{{dt}^{\mathrm{3}} } \\ $$$$\:\:\:\:\:\:\:\:+\frac{\mathrm{4}}{{t}^{\mathrm{4}} }\mathrm{sin}\:{u}\left(\frac{{du}}{{dt}}\right)^{\mathrm{2}} −\frac{\mathrm{4}}{\mathrm{3}{t}^{\mathrm{3}} }\mathrm{cos}\:{u}\left(\frac{{du}}{{dt}}\right)^{\mathrm{3}} −\frac{\mathrm{4}}{\mathrm{3}{t}^{\mathrm{3}} }\mathrm{sin}\:{u}\left(\mathrm{2}\frac{{du}}{{dt}}\right)\left(\frac{{d}^{\mathrm{2}} {u}}{{dt}^{\mathrm{2}} }\right) \\ $$$${we}\:{get}\:{an}\:{equation}\:{between}\:{t}\:{and}\:{u}. \\ $$$${still}\:{a}\:{very}\:{difficult}\:{equation} \\ $$

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