Question Number 207317 by mr W last updated on 11/May/24 $${solve}\:{for}\:{y} \\ $$$$\frac{\mathrm{1}}{{y}'}+\frac{\mathrm{1}}{{y}''}=\mathrm{1} \\ $$ Answered by Berbere last updated on 11/May/24 $${y}'={z} \\ $$$$\frac{\mathrm{1}}{{z}}+\frac{\mathrm{1}}{{z}'}=\mathrm{1}\Rightarrow{z}'=\frac{{z}}{{z}−\mathrm{1}}\Rightarrow\left(\frac{{z}−\mathrm{1}}{{z}}\right){dz}={dx}…
Question Number 206616 by universe last updated on 20/Apr/24 Answered by aleks041103 last updated on 21/Apr/24 $${e}^{\mathrm{3}{x}} \:{is}\:{increasing}\:{and}\:{continuous}\:{for}\:{x}>\mathrm{0} \\ $$$${ln}\left({x}\right)\:{is}\:{incr}.\:{and}\:{cont}.\:{for}\:{x}>\mathrm{0} \\ $$$$\Rightarrow{f}\left({x}\right)\:{is}\:{incr}.\:{and}\:{cont}.\:{for}\:{x}>\mathrm{0} \\ $$$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\:{f}\left({x}\right)\rightarrow−\infty…
Question Number 206449 by necx122 last updated on 15/Apr/24 $${solve}\:{the}\:{first}\:{order}\:{differential} \\ $$$${equation}: \\ $$$$ \\ $$$${xdy}\:−\:{ydx}\:=\:\left({xy}\right)^{\mathrm{1}/\mathrm{2}} {dx} \\ $$ Answered by Berbere last updated on…
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} \\…