Question Number 181330 by Mastermind last updated on 24/Nov/22
$$\mathrm{Solve}: \\ $$$$\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{2x}\left(\mathrm{y}+\mathrm{1}\right)=\mathrm{0},\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{y}\left(\mathrm{0}\right)=\mathrm{2} \\ $$
Answered by FelipeLz last updated on 24/Nov/22
![(dy/dx)+2x(y+1) = 0 (dy/dx)+2xy = −2x (dy/dx)e^x^2 +2xe^x^2 y = −2xe^x^2 (d/dx)[ye^x^2 ] = −2xe^x^2 ye^x^2 = −∫2xe^x^2 dx ye^x^2 = −e^x^2 +c y(x) = ce^(−x^2 ) −1 y(0) = 2 → ce^(−0^2 ) −1 = 2 ⇒ c = 3 y(x) = 3e^(−x^2 ) −1 • (dy/dx)+2x(y+1) = 0 (dy/dx) = −2x(y+1) (1/(y+1))dy = −2xdx ∫(1/(y+1))dy = −∫2xdx ln∣y+1∣+c_1 = −x^2 +c_2 ln∣y+1∣ = −x^2 +c_2 −c_1 c_2 −c_1 = c_3 → ln∣y+1∣ = −x^2 +c_3 y+1 = e^(−x^2 +c_3 ) e^c_3 = C → y+1 = Ce^(−x^2 ) y(x) = Ce^(−x^2 ) −1 y(0) = 2 → Ce^(−0^2 ) −1 = 2 ⇒ C = 3 y(x) = 3e^(−x^2 ) −1](Q181353.png)
$$\frac{{dy}}{{dx}}+\mathrm{2}{x}\left({y}+\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}+\mathrm{2}{xy}\:=\:−\mathrm{2}{x} \\ $$$$\frac{{dy}}{{dx}}{e}^{{x}^{\mathrm{2}} } +\mathrm{2}{xe}^{{x}^{\mathrm{2}} } {y}\:=\:−\mathrm{2}{xe}^{{x}^{\mathrm{2}} } \\ $$$$\frac{{d}}{{dx}}\left[{ye}^{{x}^{\mathrm{2}} } \right]\:=\:−\mathrm{2}{xe}^{{x}^{\mathrm{2}} } \\ $$$${ye}^{{x}^{\mathrm{2}} } =\:−\int\mathrm{2}{xe}^{{x}^{\mathrm{2}} } {dx} \\ $$$${ye}^{{x}^{\mathrm{2}} } \:=\:−{e}^{{x}^{\mathrm{2}} } +{c} \\ $$$${y}\left({x}\right)\:=\:{ce}^{−{x}^{\mathrm{2}} } −\mathrm{1} \\ $$$${y}\left(\mathrm{0}\right)\:=\:\mathrm{2}\:\rightarrow\:{ce}^{−\mathrm{0}^{\mathrm{2}} } −\mathrm{1}\:=\:\mathrm{2}\:\Rightarrow\:{c}\:=\:\mathrm{3} \\ $$$${y}\left({x}\right)\:=\:\mathrm{3}{e}^{−{x}^{\mathrm{2}} } −\mathrm{1} \\ $$$$\bullet \\ $$$$\frac{{dy}}{{dx}}+\mathrm{2}{x}\left({y}+\mathrm{1}\right)\:=\:\mathrm{0} \\ $$$$\frac{{dy}}{{dx}}\:=\:−\mathrm{2}{x}\left({y}+\mathrm{1}\right) \\ $$$$\frac{\mathrm{1}}{{y}+\mathrm{1}}{dy}\:=\:−\mathrm{2}{xdx} \\ $$$$\int\frac{\mathrm{1}}{{y}+\mathrm{1}}{dy}\:=\:−\int\mathrm{2}{xdx} \\ $$$$\mathrm{ln}\mid{y}+\mathrm{1}\mid+{c}_{\mathrm{1}} \:=\:−{x}^{\mathrm{2}} +{c}_{\mathrm{2}} \\ $$$$\mathrm{ln}\mid{y}+\mathrm{1}\mid\:=\:−{x}^{\mathrm{2}} +{c}_{\mathrm{2}} −{c}_{\mathrm{1}} \\ $$$${c}_{\mathrm{2}} −{c}_{\mathrm{1}} \:=\:{c}_{\mathrm{3}} \:\rightarrow\:\mathrm{ln}\mid{y}+\mathrm{1}\mid\:=\:−{x}^{\mathrm{2}} +{c}_{\mathrm{3}} \\ $$$${y}+\mathrm{1}\:=\:{e}^{−{x}^{\mathrm{2}} +{c}_{\mathrm{3}} } \\ $$$${e}^{{c}_{\mathrm{3}} } \:=\:{C}\:\rightarrow\:{y}+\mathrm{1}\:=\:{Ce}^{−{x}^{\mathrm{2}} } \\ $$$${y}\left({x}\right)\:=\:{Ce}^{−{x}^{\mathrm{2}} } −\mathrm{1} \\ $$$${y}\left(\mathrm{0}\right)\:=\:\mathrm{2}\:\rightarrow\:{Ce}^{−\mathrm{0}^{\mathrm{2}} } −\mathrm{1}\:=\:\mathrm{2}\:\Rightarrow\:{C}\:=\:\mathrm{3} \\ $$$${y}\left({x}\right)\:=\:\mathrm{3}{e}^{−{x}^{\mathrm{2}} } −\mathrm{1} \\ $$