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Category: Differential Equation

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Question Number 208569 by pticantor last updated on 18/Jun/24 $$\boldsymbol{{help}}\:\boldsymbol{{me}}\:\boldsymbol{{to}}\:\boldsymbol{{solve}}\:\boldsymbol{{this}}\:\boldsymbol{{please}} \\ $$$$\:\:\boldsymbol{{y}}''−\sqrt{\mathrm{1}+\boldsymbol{{y}}'^{\mathrm{2}} }=\boldsymbol{{x}}^{\mathrm{2}} \\ $$$$\boldsymbol{{solve}}\:\boldsymbol{{this}}\:\boldsymbol{{differential}}\:\boldsymbol{{equation}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$ Terms of…

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Question Number 207991 by efronzo1 last updated on 02/Jun/24 $$\:\:\:\:\:\mathrm{y}\:\underline{\underbrace{\mathcal{W}}} \\ $$ Answered by mr W last updated on 02/Jun/24 $${particular}\:{solution}: \\ $$$${y}={A}\:\mathrm{sin}\:{x}+{B}\:\mathrm{cos}\:{x} \\ $$$${y}'={A}\:\mathrm{cos}\:{x}−{B}\:\mathrm{sin}\:{x}…

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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-206616

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…

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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…