Question Number 217797 by Tawa11 last updated on 21/Mar/25 $$\mathrm{Solve}: \\ $$$$\:\:\:\:\:\mathrm{5x}^{\mathrm{2}} \:\mathrm{y}''\:\:+\:\:\:\mathrm{x}\left(\mathrm{1}\:\:+\:\:\mathrm{x}\right)\:\mathrm{y}'\:\:−\:\:\mathrm{y}\:\:\:=\:\:\:\mathrm{0} \\ $$ Answered by AntonCWX8 last updated on 22/Mar/25 $${I}.{F},\:\mu\left({x}\right)=\frac{\mathrm{1}}{\mathrm{5}{x}^{\mathrm{2}} }{e}^{\int\frac{{x}\left(\mathrm{1}+{x}\right)}{\mathrm{5}{x}^{\mathrm{2}} }{dx}}…
Question Number 217245 by OmoloyeMichael last updated on 07/Mar/25 $${find}\:{the}\:{following}\:{differential}\:{equation}\: \\ $$$${by}\:{eliminating}\:{the}\:{arbritrary}\:{constant} \\ $$$$\left(\mathrm{1}\right){y}={Ae}^{{x}} +{Bcosx} \\ $$$$\left(\mathrm{2}\right)\:{xy}={Ae}^{{x}} +{Be}^{−{x}} +{x}^{\mathrm{2}} \\ $$$$ \\ $$ Answered by…
Question Number 217046 by OmoloyeMichael last updated on 27/Feb/25 $${form}\:{the}\:{differential}\:{equation}\:{by}\: \\ $$$${eliminating}\:{the}\:{arbritrary}\:{constant} \\ $$$${y}^{\mathrm{2}} ={Ax}^{\mathrm{2}} +{Bx}+{C} \\ $$ Answered by mr W last updated on…
Question Number 216982 by OmoloyeMichael last updated on 26/Feb/25 $$\boldsymbol{{Form}}\:\boldsymbol{{the}}\:\boldsymbol{{differential}}\:\boldsymbol{{equation}}\:\boldsymbol{{by}} \\ $$$$\boldsymbol{{eliminating}}\:\boldsymbol{{abitrary}}\:\boldsymbol{{constant}}. \\ $$$$\boldsymbol{{y}}^{\mathrm{2}} =\boldsymbol{{Ax}}^{\mathrm{2}} +\boldsymbol{{Bx}}+\boldsymbol{{C}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 216787 by Engr_Jidda last updated on 20/Feb/25 $${form}\:{the}\:{differential}\:{equationfrom}\:{the}\:{following} \\ $$$$\left.\mathrm{1}\right)\:{y}={Ae}^{\mathrm{3}{x}} +{Be}^{\mathrm{5}{x}} \\ $$$$\left.\mathrm{2}\right)\:{y}^{\mathrm{2}} =\left({x}−\mathrm{1}\right) \\ $$$$\left.\mathrm{3}\right)\:{c}\left({y}+{c}\right)^{\mathrm{2}} +{x}^{\mathrm{3}} =\mathrm{0} \\ $$ Answered by som(math1967)…
Question Number 215468 by alephnull last updated on 07/Jan/25 $$\frac{\partial{x}\omega}{\partial{y}\omega}\centerdot\frac{\partial{e}}{\partial\omega}=? \\ $$$${x}={f}\left({y}\right) \\ $$$$\omega={g}\left({y}\right) \\ $$$${e}={h}\left(\omega\right) \\ $$ Answered by MrGaster last updated on 08/Jan/25…
Question Number 214813 by ajfour last updated on 20/Dec/24 $${Help}\:{me}\:{solve}\:{this} \\ $$$$\frac{{dy}}{{dx}}+\frac{{a}}{{y}}+{b}\sqrt{{x}}=\mathrm{0} \\ $$ Commented by ajfour last updated on 20/Dec/24 https://youtu.be/H53vv0XMRs4?si=eeFguqQqwBJDXTTU Answered by TonyCWX08…
Question Number 213371 by otchoumou last updated on 03/Nov/24 Commented by otchoumou last updated on 03/Nov/24 $${aider} \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 213097 by issac last updated on 30/Oct/24 $$\mathrm{Uhhhh}. \\ $$$$\mathrm{can}\:\mathrm{you}\:\mathrm{guys}\:\mathrm{solve}\:\mathrm{Partial}\:\mathrm{differantial}\:\mathrm{equation} \\ $$$$\bigtriangledown^{\mathrm{2}} \boldsymbol{\phi}=\mathrm{0} \\ $$$$\mathrm{Cylinderical}\:\mathrm{Laplacian}\:\mathrm{case} \\ $$$$\bigtriangledown^{\mathrm{2}} =\frac{\mathrm{1}}{\rho}\centerdot\frac{\partial\:\:}{\partial\rho}\left(\rho\frac{\partial\:\:}{\partial\rho}\right)+\left(\frac{\mathrm{1}}{\rho}\right)^{\mathrm{2}} \frac{\partial^{\mathrm{2}} \:}{\partial\phi^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} \:\:}{\partial{z}^{\mathrm{2}} }…
Question Number 212895 by papuchin last updated on 26/Oct/24 Terms of Service Privacy Policy Contact: info@tinkutara.com