Question Number 201555 by Simurdiera last updated on 08/Dec/23 Answered by mr W last updated on 09/Dec/23 $${let}\:{u}={x}+{y} \\ $$$$\frac{{du}}{{dx}}=\mathrm{1}+\frac{{dy}}{{dx}}\:\Rightarrow\frac{{dy}}{{dx}}=\frac{{du}}{{dx}}−\mathrm{1} \\ $$$$\Rightarrow\sqrt{{u}+\mathrm{1}}\left(\frac{{du}}{{dx}}−\mathrm{1}\right)=\sqrt{{u}−\mathrm{1}} \\ $$$$\Rightarrow\frac{{du}}{{dx}}=\frac{\sqrt{{u}−\mathrm{1}}}{\:\sqrt{{u}+\mathrm{1}}}+\mathrm{1} \\…
Question Number 201421 by mr W last updated on 06/Dec/23 Commented by mr W last updated on 06/Dec/23 Commented by mr W last updated on…
Question Number 200902 by Rupesh123 last updated on 26/Nov/23 Answered by Rasheed.Sindhi last updated on 26/Nov/23 $${f}\left({x}\right){f}\left({y}\right)+\mathrm{1}=\mathrm{2}{f}\left({xy}\right)+\mathrm{2}\left({x}+{y}\right) \\ $$$${x}={y}=\mathrm{1}: \\ $$$$\left[{f}\left(\mathrm{1}\right)\right]^{\mathrm{2}} +\mathrm{1}=\mathrm{2}{f}\left(\mathrm{1}\right)+\mathrm{4} \\ $$$$\left[{f}\left(\mathrm{1}\right)\right]^{\mathrm{2}} −\mathrm{2}{f}\left(\mathrm{1}\right)−\mathrm{3}=\mathrm{0}…
Question Number 200596 by Rupesh123 last updated on 20/Nov/23 Answered by witcher3 last updated on 22/Nov/23 $$\mathrm{a}\:\mathrm{True}\: \\ $$$$\mathrm{b}\:\mathrm{we}\:\mathrm{can}\:\mathrm{show}\:\mathrm{that}\:\mathrm{exist}\:\mathrm{bijection}\:\mathrm{between}\:\mathrm{som}\:\:\mathrm{set}\:\mathrm{of}\:\mathrm{not}\:\mathrm{differentiabl} \\ $$$$\mathrm{point}\:\mathrm{and}\:\mathbb{N}\: \\ $$$$ \\ $$…
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Question Number 199996 by jlewis last updated on 12/Nov/23 $$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{first}\:\mathrm{order}\:\mathrm{energy}\:\mathrm{correction}\:\mathrm{for} \\ $$$$\mathrm{1}−\mathrm{dimensional}\:\mathrm{non}−\mathrm{degenerate}\:\mathrm{anharmonic} \\ $$$$\mathrm{oscillator}\:\mathrm{whose}\:\mathrm{harmiltonian}\:\mathrm{is}\:\mathscr{H}\underline{\mathscr{L}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 200022 by jlewis last updated on 12/Nov/23 $$\mathrm{solve}\:\mathrm{the}\:\mathrm{associated}\:\mathrm{legendre}\:\mathrm{equation} \\ $$$$\lambda={l}\:\left({l}+\mathrm{1}\right)\eta^{\mathrm{2}} \:;{l}=\mathrm{0},\mathrm{1},\mathrm{2}…\:\:\:{and}\:{m}^{\mathrm{2}} \leqslant\:{l}\left({l}+\mathrm{1}\right)\: \\ $$$${which}\:{requires}\:−{l}\leqslant{m}\leqslant{l}\:\mathrm{using}\:\mathrm{power}\:\mathrm{series} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 199844 by jlewis last updated on 10/Nov/23 $$\mathrm{solve}\:\frac{\mathrm{d}^{\mathrm{2}} }{\mathrm{dx}^{\mathrm{2}} }\:\mathrm{x} \\ $$ Answered by AST last updated on 10/Nov/23 $$\frac{{d}}{{dx}}\left(\frac{{d}}{{dx}}{x}\right)=\frac{{d}}{{dx}}\left(\mathrm{1}\right)=\mathrm{0} \\ $$ Answered…
Question Number 199891 by Mathspace last updated on 10/Nov/23 $${solve}\:{by}\:{laplce}\:{transform} \\ $$$${y}^{''} −{y}^{'} +{y}\:=\left({x}+\mathrm{1}\right){e}^{{x}} \\ $$ Commented by Mathspace last updated on 10/Nov/23 $${with}\:{y}\left(\mathrm{0}\right)=\mathrm{1}\:{and}\:{y}^{'} \left(\mathrm{0}\right)=−\mathrm{1}…
Question Number 199876 by aurpeyz last updated on 10/Nov/23 Terms of Service Privacy Policy Contact: info@tinkutara.com