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Category: Differential Equation

Solve-1-x-dy-dx-y-1-x-

Question Number 15095 by tawa tawa last updated on 07/Jun/17 $$\mathrm{Solve}: \\ $$$$\left(\mathrm{1}\:−\:\mathrm{x}\right)\frac{\mathrm{dy}}{\mathrm{dx}}\:=\:\mathrm{y}\left(\mathrm{1}\:+\:\mathrm{x}\right) \\ $$ Answered by Tinkutara last updated on 07/Jun/17 $$\frac{{dy}}{{y}}\:=\:\frac{\mathrm{1}\:+\:{x}}{\mathrm{1}\:−\:{x}}\:{dx}\:=\:\left(−\mathrm{1}\:−\:\frac{\mathrm{2}}{{x}\:−\:\mathrm{1}}\right){dx} \\ $$$$\mathrm{ln}\:{y}\:=\:−{x}\:−\:\mathrm{2}\:\mathrm{ln}\:\mid{x}\:−\:\mathrm{1}\mid\:+\:{C}…

Question-146057

Question Number 146057 by mnjuly1970 last updated on 10/Jul/21 Answered by mathmax by abdo last updated on 10/Jul/21 $$\mathrm{J}=\mathrm{Im}\left(\int_{\mathrm{0}} ^{\infty} \:\mathrm{x}^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\mathrm{e}^{−\mathrm{x}+\mathrm{ix}} \mathrm{dx}\right)\:\:\mathrm{we}\:\mathrm{have} \\ $$$$\int_{\mathrm{0}}…

An-open-box-of-area-486cm-2-If-the-length-is-twice-the-breadth-Find-the-maximum-volume-of-the-box-hence-Show-the-volume-is-maximum-

Question Number 14543 by chux last updated on 02/Jun/17 $$\mathrm{An}\:\mathrm{open}\:\mathrm{box}\:\mathrm{of}\:\mathrm{area}\:\mathrm{486cm}^{\mathrm{2}} .\mathrm{If}\:\mathrm{the} \\ $$$$\mathrm{length}\:\mathrm{is}\:\mathrm{twice}\:\mathrm{the}\:\mathrm{breadth}.\mathrm{Find} \\ $$$$\mathrm{the}\:\mathrm{maximum}\:\mathrm{volume}\:\mathrm{of}\:\mathrm{the}\:\mathrm{box}. \\ $$$$\mathrm{hence},\mathrm{Show}\:\mathrm{the}\:\mathrm{volume}\:\mathrm{is}\:\mathrm{maximum}. \\ $$ Commented by chux last updated on…

Solve-the-differential-equation-y-2x-3y-4-4x-3y-2-

Question Number 14444 by tawa tawa last updated on 31/May/17 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation}\: \\ $$$$\mathrm{y}'\:=\:\frac{\mathrm{2x}\:+\:\mathrm{3y}\:−\:\mathrm{4}}{\mathrm{4x}\:+\:\mathrm{3y}\:+\:\mathrm{2}} \\ $$ Answered by mrW1 last updated on 31/May/17 $${let}\:{x}={u}+{p}\:{and}\:{y}={v}+{q} \\ $$$$\mathrm{2}{x}+\mathrm{3}{y}−\mathrm{4}=\mathrm{2}{u}+\mathrm{3}{v}+\mathrm{2}{p}+\mathrm{3}{q}−\mathrm{4}…