Question Number 223988 by MirHasibulHossain last updated on 12/Aug/25 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{DE}\:\mathrm{using}\:\mathrm{the}\:\mathrm{method}\:\mathrm{of}\:\mathrm{Frobenius}\::\: \\ $$$$\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)\mathrm{y}''−\mathrm{2xy}'+\mathrm{n}\left(\mathrm{n}+\mathrm{1}\right)\mathrm{y}=\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 223054 by Silver last updated on 13/Jul/25 $$\mathrm{find}\:{y} \\ $$$${y}^{{dy}} =\:{x}^{{dx}} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 222296 by Shrodinger last updated on 22/Jun/25 $$\left(\mathrm{1}+{x}^{\mathrm{4}} \right){y}'−{x}^{\mathrm{3}} {y}\:=\:{x}^{\mathrm{5}} −{x}^{\mathrm{3}} +\mathrm{2}{x}+\mathrm{1} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 221973 by ajfour last updated on 14/Jun/25 $$\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+{y}={k}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{6}}{{x}^{\mathrm{4}} }\:\:\:\:\:\: \\ $$$${Find}\:{y}\left({x}\right)\:\:\:\:\left({k}\:{is}\:{constant}\right). \\ $$ Answered by mahdipoor last updated on 14/Jun/25…
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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)…