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Question Number 133570 by Ahmed1hamouda last updated on 23/Feb/21

Commented by Ahmed1hamouda last updated on 23/Feb/21

Solve the differential equations

Commented by EDWIN88 last updated on 23/Feb/21

fragst du ernsthaft

$$ \\ $$$$\mathrm{fragst}\:\mathrm{du}\:\mathrm{ernsthaft} \\ $$

Answered by EDWIN88 last updated on 23/Feb/21

Solve the differential equation (dy/dx)+ ((3x+2y−5)/(2x+3y−5)) = 0 set { ((x=x_1 +h)),((y=y_1 +k)) :} (dy_1 /dx_1 ) = ((−3x_1 −2y_1 −3h−2k+5)/(2x_1 +3y_1 +2h+3k+5)) solving set of two equation { ((−3h−2k+5=0)),((2h+3k+5=0)) :} we find h=1 and k=1 the result (dy_1 /dx_1 ) = ((−3x_1 −2y_1 )/(2x_1 +3y_1 )) ; let y_1 =ux_1 ⇒ u + x_1 (du/dx_1 ) = ((−3x_1 −2ux_1 )/(2x_1 +3ux_1 )) = ((−3−2u)/(2+3u)) ⇒ x_1 (du/dx_1 ) = ((−3−2u−2u−3u^2 )/(2+3u)) ⇒ (((2+3u))/(3u^2 +4u+3))du = −(dx_1 /x_1 ) ⇒∫ ((d(3u^2 +4u+3))/(3u^2 +4u+3))=−2∫ (dx_1 /x_1 ) ln ∣3u^2 +4u+3∣ = ln ∣(1/x_1 ^2 ) ∣+c ⇒ 3u^2 +4u+3 = (C/x_1 ^2 ) ⇒3((y_1 /x_1 ))^2 +4((y_1 /x_1 ))+3 = (C/x_1 ^2 ) ⇒3y_1 ^2 +4x_1 y_1 +3x_1 ^2 = C ⇒3(y−1)^2 +4(x−1)(y−1)+3(x−1)^2 =C

$$\:\mathrm{Solve}\:\mathrm{the}\:\mathrm{differential}\:\mathrm{equation}\: \\ $$$$\:\frac{\mathrm{dy}}{\mathrm{dx}}+\:\frac{\mathrm{3x}+\mathrm{2y}−\mathrm{5}}{\mathrm{2x}+\mathrm{3y}−\mathrm{5}}\:=\:\mathrm{0}\: \\ $$$$\mathrm{set}\:\begin{cases}{\mathrm{x}=\mathrm{x}_{\mathrm{1}} +\mathrm{h}}\\{\mathrm{y}=\mathrm{y}_{\mathrm{1}} +\mathrm{k}}\end{cases} \\ $$$$\frac{\mathrm{dy}_{\mathrm{1}} }{\mathrm{dx}_{\mathrm{1}} }\:=\:\frac{−\mathrm{3x}_{\mathrm{1}} −\mathrm{2y}_{\mathrm{1}} −\mathrm{3h}−\mathrm{2k}+\mathrm{5}}{\mathrm{2x}_{\mathrm{1}} +\mathrm{3y}_{\mathrm{1}} +\mathrm{2h}+\mathrm{3k}+\mathrm{5}} \\ $$$$\mathrm{solving}\:\mathrm{set}\:\mathrm{of}\:\mathrm{two}\:\mathrm{equation}\:\begin{cases}{−\mathrm{3h}−\mathrm{2k}+\mathrm{5}=\mathrm{0}}\\{\mathrm{2h}+\mathrm{3k}+\mathrm{5}=\mathrm{0}}\end{cases} \\ $$$$\mathrm{we}\:\mathrm{find}\:\mathrm{h}=\mathrm{1}\:\mathrm{and}\:\mathrm{k}=\mathrm{1}\: \\ $$$$\mathrm{the}\:\mathrm{result}\:\frac{\mathrm{dy}_{\mathrm{1}} }{\mathrm{dx}_{\mathrm{1}} }\:=\:\frac{−\mathrm{3x}_{\mathrm{1}} −\mathrm{2y}_{\mathrm{1}} }{\mathrm{2x}_{\mathrm{1}} +\mathrm{3y}_{\mathrm{1}} }\:;\:\mathrm{let}\:\mathrm{y}_{\mathrm{1}} =\mathrm{ux}_{\mathrm{1}} \\ $$$$\Rightarrow\:\mathrm{u}\:+\:\mathrm{x}_{\mathrm{1}} \:\frac{\mathrm{du}}{\mathrm{dx}_{\mathrm{1}} }\:=\:\frac{−\mathrm{3x}_{\mathrm{1}} −\mathrm{2ux}_{\mathrm{1}} }{\mathrm{2x}_{\mathrm{1}} +\mathrm{3ux}_{\mathrm{1}} }\:=\:\frac{−\mathrm{3}−\mathrm{2u}}{\mathrm{2}+\mathrm{3u}} \\ $$$$\Rightarrow\:\mathrm{x}_{\mathrm{1}} \:\frac{\mathrm{du}}{\mathrm{dx}_{\mathrm{1}} }\:=\:\frac{−\mathrm{3}−\mathrm{2u}−\mathrm{2u}−\mathrm{3u}^{\mathrm{2}} }{\mathrm{2}+\mathrm{3u}} \\ $$$$\Rightarrow\:\frac{\left(\mathrm{2}+\mathrm{3u}\right)}{\mathrm{3u}^{\mathrm{2}} +\mathrm{4u}+\mathrm{3}}\mathrm{du}\:=\:−\frac{\mathrm{dx}_{\mathrm{1}} }{\mathrm{x}_{\mathrm{1}} } \\ $$$$\Rightarrow\int\:\frac{\mathrm{d}\left(\mathrm{3u}^{\mathrm{2}} +\mathrm{4u}+\mathrm{3}\right)}{\mathrm{3u}^{\mathrm{2}} +\mathrm{4u}+\mathrm{3}}=−\mathrm{2}\int\:\frac{\mathrm{dx}_{\mathrm{1}} }{\mathrm{x}_{\mathrm{1}} } \\ $$$$\mathrm{ln}\:\mid\mathrm{3u}^{\mathrm{2}} +\mathrm{4u}+\mathrm{3}\mid\:=\:\mathrm{ln}\:\mid\frac{\mathrm{1}}{\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} }\:\mid+\mathrm{c} \\ $$$$\Rightarrow\:\mathrm{3u}^{\mathrm{2}} +\mathrm{4u}+\mathrm{3}\:=\:\frac{\mathrm{C}}{\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{3}\left(\frac{\mathrm{y}_{\mathrm{1}} }{\mathrm{x}_{\mathrm{1}} }\right)^{\mathrm{2}} +\mathrm{4}\left(\frac{\mathrm{y}_{\mathrm{1}} }{\mathrm{x}_{\mathrm{1}} }\right)+\mathrm{3}\:=\:\frac{\mathrm{C}}{\mathrm{x}_{\mathrm{1}} ^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{3y}_{\mathrm{1}} ^{\mathrm{2}} \:+\mathrm{4x}_{\mathrm{1}} \mathrm{y}_{\mathrm{1}} +\mathrm{3x}_{\mathrm{1}} ^{\mathrm{2}} \:=\:\mathrm{C} \\ $$$$\Rightarrow\mathrm{3}\left(\mathrm{y}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}\left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{y}−\mathrm{1}\right)+\mathrm{3}\left(\mathrm{x}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{C} \\ $$

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