Menu Close

Category: Differential Equation

Question-127354

Question Number 127354 by liberty last updated on 29/Dec/20 Answered by som(math1967) last updated on 29/Dec/20 $$\mathrm{1}.\:{let}\:{x}={X}+{h}\:\:\:{y}={Y}+{k} \\ $$$$\frac{{dx}}{{dX}}=\mathrm{1}\:\:\:\:\frac{{dy}}{{dY}}=\mathrm{1} \\ $$$$\frac{{dy}}{{dx}}=\frac{{dy}}{{dY}}\:.\frac{{dY}}{{dX}}.\frac{{dX}}{{dx}}=\mathrm{1}.\frac{{dY}}{{dX}}.\mathrm{1}=\frac{{dY}}{{dX}} \\ $$$$\therefore\frac{{dY}}{{dX}}=\frac{{X}+{h}+{Y}+{k}−\mathrm{3}}{{X}+{h}−{Y}−{k}−\mathrm{1}} \\ $$$$\frac{{dY}}{{dX}}=\frac{{X}+{Y}+{h}+{k}−\mathrm{3}}{{X}−{Y}+{h}−{k}−\mathrm{1}}…

x-2-1-y-2xy-2y-x-2-1-2-

Question Number 127260 by bramlexs22 last updated on 28/Dec/20 $$\left({x}^{\mathrm{2}} −\mathrm{1}\right){y}''−\mathrm{2}{xy}'+\mathrm{2}{y}\:=\:\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} \\ $$ Answered by liberty last updated on 28/Dec/20 $${let}\:{y}\:=\:{xz}\:{for}\:{some}\:{z}={z}\left({x}\right) \\ $$$${y}'={xz}'+{z}\:\wedge\:{y}''={xz}''+\mathrm{2}{z}' \\…

2-0-x-y-3-cos-xdx-yd-2-y-dx-2-dy-dx-2-ky-5-sin-x-y-0-a-y-0-0-solve-the-differential-equation-Laplace-tranforms-might-be-helpful-i-think-

Question Number 61635 by ajfour last updated on 05/Jun/19 $$\mathrm{2}\left(\int_{\mathrm{0}} ^{\:{x}} {y}^{\mathrm{3}} \mathrm{cos}\:{xdx}\right)\left[\frac{{yd}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }−\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} \right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:=\:{ky}^{\mathrm{5}} \mathrm{sin}\:{x}\:\:\:\:\:;\:\: \\ $$$$\:\:{y}\left(\mathrm{0}\right)={a},\:{y}'\left(\mathrm{0}\right)=\mathrm{0}\:. \\ $$$$\:{solve}\:{the}\:{differential}\:{equation}. \\ $$$$\left({Laplace}\:{tranforms}\:{might}\right.…

D-2-1-y-x-sin-x-

Question Number 127064 by benjo_mathlover last updated on 26/Dec/20 $$\:\:\:\left({D}^{\mathrm{2}} −\mathrm{1}\right){y}\:=\:{x}\:\mathrm{sin}\:{x}\: \\ $$ Answered by liberty last updated on 26/Dec/20 $$\:{The}\:{characteristic}\:{eq}\:{of}\:\left({D}^{\mathrm{2}} −\mathrm{1}\right){y}\:=\:\mathrm{0} \\ $$$${is}\:\lambda^{\mathrm{2}} −\mathrm{1}=\mathrm{0}\:,\:{has}\:{the}\:{roots}\:\lambda=\pm\mathrm{1}…

Question-192345

Question Number 192345 by Mingma last updated on 15/May/23 Answered by aleks041103 last updated on 15/May/23 $$\int_{\mathrm{0}} ^{\:\mathrm{1}} \left({f}\left(\mathrm{1}+{x}\right)+{f}\left(\mathrm{1}−{x}\right)\right){dx}= \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {f}\left(\mathrm{1}+{x}\right){dx}\:−\:\int_{\mathrm{0}} ^{\:\mathrm{1}} {f}\left(\mathrm{1}−{x}\right){d}\left(−{x}\right)=…