Question Number 206142 by universe last updated on 07/Apr/24 $$\:\:\:\:\mathrm{let}\:\:\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+{p}\left({x}\right)\frac{{dy}}{{dx}}+{q}\left({x}\right){y}=\mathrm{0}\:,\:{x}\in\mathbb{R}\:\mathrm{where}\: \\ $$$$\:\:\:\:{p}\left({x}\right)\:\mathrm{and}\:{q}\left({x}\right)\:\mathrm{are}\:\mathrm{continuous}\:\mathrm{function}\:\mathrm{if} \\ $$$$\:\:\:\:{y}_{\mathrm{1}} =\:\mathrm{sin}{x}−\mathrm{2cos}{x}\:{and}\:{y}_{\mathrm{2}} \:=\:\mathrm{2sin}{x}\:+\mathrm{cos}{x} \\ $$$$\:\:\:\:\mathrm{are}\:{L}.{I}\:\left(\mathrm{linearly}\:\mathrm{independent}\right)\:\mathrm{solution} \\ $$$$\:\:\:\:\:\mathrm{then}\:\:\mid\mathrm{4}{p}\left(\mathrm{0}\right)+\mathrm{2}{q}\left(\mathrm{1}\right)\mid\:=\:?\:\:\: \\ $$ Answered…
Question Number 206047 by universe last updated on 05/Apr/24 Answered by aleks041103 last updated on 06/Apr/24 $${y}'={y}+{const}. \\ $$$$\Rightarrow{y}={c}_{\mathrm{2}} {e}^{{x}} −{c}_{\mathrm{1}} \\ $$$$\Rightarrow{y}'={y}+{c}_{\mathrm{1}} \\ $$$$\Rightarrow{c}_{\mathrm{1}}…
Question Number 205184 by Simurdiera last updated on 12/Mar/24 $${Resuelve}\:{la}\:{siguiente}\:{ecuaci}\acute {{o}n}\:{diferencial} \\ $$$$\frac{{dx}}{{dy}}\:+\:{x}^{\mathrm{2}} \:=\:\frac{\mathrm{1}}{{y}^{\mathrm{4}} } \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 203898 by necx122 last updated on 02/Feb/24 $${Let}\:{f}\left({W}\right)\:{be}\:{a}\:{function}\:{of}\:{vector}\:{W}\:\in\: {R}^{{N}} , \\ $$$${i}.{e}.\:{f}\left({W}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}\:+\:{e}^{−{W}^{{T}} {x}} } \\ $$$${Determine}\:{the}\:{first}\:{derivative}\:{and} \\ $$$${matrix}\:{of}\:{second}\:{derivatives}\:{of}\:{f}\:{with} \\ $$$${respect}\:{to}\:{W} \\ $$$$ \\…
Question Number 203694 by Numsey last updated on 26/Jan/24 Answered by Calculusboy last updated on 26/Jan/24 $$\boldsymbol{{Solution}}:\:\boldsymbol{{by}}\:\boldsymbol{{sub}}\:\boldsymbol{{directly}},\boldsymbol{{we}}\:\boldsymbol{{get}}\:\frac{\mathrm{0}}{\mathrm{0}}\left(\boldsymbol{{indeterminant}}\right) \\ $$$$\boldsymbol{{let}}\:\boldsymbol{\Delta}=\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\left(\mathrm{1}+\boldsymbol{{x}}\right)^{\mathrm{5}} }{\boldsymbol{{x}}^{\mathrm{2}} }−\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}} {\boldsymbol{{m}}}\frac{\boldsymbol{{e}}^{\mathrm{5}\boldsymbol{{x}}} }{\boldsymbol{{x}}^{\mathrm{2}} }+\boldsymbol{{li}}\underset{\boldsymbol{{x}}\rightarrow\mathrm{0}}…
Question Number 203498 by ajfour last updated on 20/Jan/24 $$\frac{{d}^{\mathrm{3}\:} {y}}{{dx}^{\mathrm{3}} }=\mathrm{4}\left({x}+\frac{\mathrm{1}}{\mathrm{4}}\right)^{\mathrm{2}} −\mathrm{4}{y} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 201555 by Simurdiera last updated on 08/Dec/23 Answered by mr W last updated on 09/Dec/23 $${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}\: \\ $$$$\mathrm{b}\:\mathrm{we}\:\mathrm{can}\:\mathrm{show}\:\mathrm{that}\:\mathrm{exist}\:\mathrm{bijection}\:\mathrm{between}\:\mathrm{som}\:\:\mathrm{set}\:\mathrm{of}\:\mathrm{not}\:\mathrm{differentiabl} \\ $$$$\mathrm{point}\:\mathrm{and}\:\mathbb{N}\: \\ $$$$ \\ $$…