Question Number 108053 by john santu last updated on 14/Aug/20 $$\:\:\:\:\:\:\:\:\:\:\:\frac{\curlywedge\mathcal{J}\:\mathbb{S}\curlywedge}{\:^{\rightrightarrows} } \\ $$$$\:\:\:\:\:\left(\frac{{y}}{{x}}+\sqrt{\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{{x}^{\mathrm{2}} }}\right){dx}={dy} \\ $$ Answered by bemath last updated on…
Question Number 108047 by bemath last updated on 14/Aug/20 $$\:\:\:\:\:\:\frac{°\bullet\mathcal{B}{e}\mathcal{M}{ath}\bullet°}{\Sigma} \\ $$$$\:\:\:{y}−{x}\:\frac{{dy}}{{dx}}\:=\:{a}\:\left(\mathrm{1}+{x}^{\mathrm{2}} \:\frac{{dy}}{{dx}}\right)\: \\ $$ Answered by john santu last updated on 14/Aug/20 $$\:\:\:\:\:\:\frac{\diamondsuit\mathcal{JS}\bigstar}{\checkmark} \\…
Question Number 108022 by Ar Brandon last updated on 13/Aug/20 $$\left(\frac{\mathrm{d}^{\mathrm{2}} }{\mathrm{dx}^{\mathrm{2}} }−\mathrm{2x}\frac{\mathrm{d}}{\mathrm{dx}}+\mathrm{2n}\right)\mathrm{H}_{\mathrm{n}} \left(\mathrm{x}\right)=\mathrm{0} \\ $$$$\mathrm{Determine}\:\mathrm{the}\:\mathrm{first}-\mathrm{4}\:\mathrm{polynomials}\:\mathrm{of}\:\mathrm{Hermite}\left(\mathrm{H}_{\mathrm{n}} \right) \\ $$ Terms of Service Privacy Policy Contact:…
Question Number 108018 by Ar Brandon last updated on 13/Aug/20 $$\mathrm{i}.\:\:\mathrm{x}^{\mathrm{3}} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}^{\mathrm{2}} +\mathrm{x}^{\mathrm{2}} \mathrm{y}+\mathrm{2x}^{\mathrm{4}} =\mathrm{0} \\ $$$$\mathrm{ii}.\:\:\frac{\mathrm{dy}}{\mathrm{dx}}=−\mathrm{2}−\mathrm{y}+\mathrm{y}^{\mathrm{2}} \\ $$$$\mathrm{iii}.\:\:\mathrm{2cos}\left(\mathrm{x}\right)\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{2cos}^{\mathrm{2}} \left(\mathrm{x}\right)−\mathrm{sin}^{\mathrm{2}} \left(\mathrm{x}\right)+\mathrm{y}^{\mathrm{2}} \:;\:\mathrm{y}\left(\mathrm{0}\right)=−\mathrm{1} \\ $$ Answered…
Question Number 108005 by Ar Brandon last updated on 13/Aug/20 $$\mathrm{1}.\:\:\mathrm{x}''\left(\mathrm{t}\right)+\mathrm{x}\left(\mathrm{t}\right)=\mathrm{tcos}\left(\mathrm{2t}\right)+\left(\mathrm{1}+\mathrm{t}^{\mathrm{2}} \right)\mathrm{sin}\left(\mathrm{2t}\right) \\ $$$$\mathrm{2}.\:\:\mathrm{x}''\left(\mathrm{t}\right)+\mathrm{x}\left(\mathrm{t}\right)=\mathrm{t}^{\mathrm{2}} \mathrm{cos}\left(\mathrm{2t}\right) \\ $$ Answered by mathmax by abdo last updated on…
Question Number 107995 by Ar Brandon last updated on 13/Aug/20 $$\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation}; \\ $$$$\mathrm{x}''\left(\mathrm{t}\right)+\mathrm{2x}'\left(\mathrm{t}\right)+\mathrm{x}\left(\mathrm{t}\right)=\mathrm{1}+\mathrm{t} \\ $$$$\left(\mathrm{using}\:\mathrm{the}\:\mathrm{method}\:\mathrm{of}\:\mathrm{variation}\:\mathrm{of}\:\mathrm{parameters}\right) \\ $$ Answered by Ar Brandon last updated on 13/Aug/20…
Question Number 107727 by Ar Brandon last updated on 12/Aug/20 $$\mathrm{Determine}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{solutions}\:\mathrm{to}\:\mathrm{the}\:\mathrm{equation}\:; \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}+\mathrm{t}^{\mathrm{3}} \right){x}'\left(\mathrm{t}\right)+\mathrm{t}^{\mathrm{2}} {x}\left(\mathrm{t}\right)+\mathrm{t}\left({x}\left(\mathrm{t}\right)\right)^{\mathrm{2}} =\mathrm{0} \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com
Question Number 107373 by Dwaipayan Shikari last updated on 10/Aug/20 $$\frac{\mathrm{d}^{\mathrm{2}} \mathrm{y}}{\mathrm{dx}^{\mathrm{2}} }=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)^{\mathrm{n}} \\ $$ Answered by hgrocks last updated on 10/Aug/20 Commented by Dwaipayan…
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Question Number 107151 by john santu last updated on 09/Aug/20 $$\:\:\:\:\:\:\:\:\:@\mathrm{JS}@ \\ $$$$\left(\mathrm{D}^{\mathrm{2}} +\mathrm{7D}+\mathrm{12}\right)\mathrm{y}\:=\:\mathrm{e}^{\mathrm{x}} \:\mathrm{cos}\:\mathrm{2x}\: \\ $$ Terms of Service Privacy Policy Contact: info@tinkutara.com