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Question Number 182069 by Mastermind last updated on 03/Dec/22

(1+x^2 )(dy/dx)+3xy=5x solve

$$\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)\frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{3xy}=\mathrm{5x} \\ $$$$ \\ $$$$\mathrm{solve} \\ $$

Commented by CElcedricjunior last updated on 04/Dec/22

(1+x^2 )(dy/dx)=x(5−3y) =>(1/(5−3y))dy=(x/(1+x^2 ))dx =>−(1/3)∫((−3)/(5−3y))dy=(1/2)∫((2x)/(1+x^2 ))dx =>−(1/3)ln∣5−3y∣=(1/2)ln∣1+x^2 ∣+c ∀x∈](3/5);+∞[ =>−(1/3)ln(5−3y)=(1/2)ln(1+x^2 )+c =>(1/( ^3 (√(5−3y))))=e^c ((√(1+x^2 ))) =>y=(1/3)(5−(e^(−3c) /( (√((1+x^2 )^3 ))))) avec c∈R y=(1/3)(5−(e^(−3c) /( (√((1+x^2 )^3 ))))) ====================== .....................cedric junior............

$$\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} \right)\frac{\boldsymbol{{dy}}}{\boldsymbol{{dx}}}=\boldsymbol{{x}}\left(\mathrm{5}−\mathrm{3}\boldsymbol{{y}}\right) \\ $$$$=>\frac{\mathrm{1}}{\mathrm{5}−\mathrm{3}\boldsymbol{{y}}}\boldsymbol{{dy}}=\frac{\boldsymbol{{x}}}{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }\boldsymbol{{dx}} \\ $$$$=>−\frac{\mathrm{1}}{\mathrm{3}}\int\frac{−\mathrm{3}}{\mathrm{5}−\mathrm{3}{y}}\boldsymbol{{dy}}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}}{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }\boldsymbol{{dx}} \\ $$$$=>−\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{ln}}\mid\mathrm{5}−\mathrm{3}\boldsymbol{{y}}\mid=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{\mathrm{ln}}\mid\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} \mid+\boldsymbol{{c}} \\ $$$$\left.\forall\boldsymbol{{x}}\in\right]\frac{\mathrm{3}}{\mathrm{5}};+\infty\left[\right. \\ $$$$=>−\frac{\mathrm{1}}{\mathrm{3}}\boldsymbol{{ln}}\left(\mathrm{5}−\mathrm{3}\boldsymbol{{y}}\right)=\frac{\mathrm{1}}{\mathrm{2}}\boldsymbol{{ln}}\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} \right)+\boldsymbol{{c}} \\ $$$$=>\frac{\mathrm{1}}{\overset{\mathrm{3}} {\:}\sqrt{\mathrm{5}−\mathrm{3}\boldsymbol{{y}}}}=\boldsymbol{{e}}^{\boldsymbol{{c}}} \left(\sqrt{\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} }\right) \\ $$$$=>\boldsymbol{{y}}=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{5}−\frac{\boldsymbol{{e}}^{−\mathrm{3}\boldsymbol{{c}}} }{\:\sqrt{\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} \right)^{\mathrm{3}} }}\right)\:\:\:\boldsymbol{{avec}}\:\boldsymbol{{c}}\in\mathbb{R} \\ $$$${y}=\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{5}−\frac{\boldsymbol{{e}}^{−\mathrm{3}\boldsymbol{{c}}} }{\:\sqrt{\left(\mathrm{1}+\boldsymbol{{x}}^{\mathrm{2}} \right)^{\mathrm{3}} }}\right) \\ $$$$====================== \\ $$$$.....................{cedric}\:{junior}............ \\ $$

Answered by Ar Brandon last updated on 03/Dec/22

(1+x^2 )(dy/dx)+3xy=5x ⇒(1+x^2 )dy=x(5−3y)dx ⇒(1/(5−3y))dy=(x/(1+x^2 ))dx ⇒∫(1/(5−3y))dy=∫(x/(1+x^2 ))dx ⇒−(1/3)ln∣5−3y∣=(1/2)ln(1+x^2 )+k ⇒5−3y=(C/( (√((1+x^2 )^3 )))) ⇒ determinant (((y=(5/3)−(C/(3(√((1+x^2 )^3 ))))))), C=e^k

$$\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\frac{{d}\mathrm{y}}{{dx}}+\mathrm{3}{x}\mathrm{y}=\mathrm{5}{x}\:\Rightarrow\left(\mathrm{1}+{x}^{\mathrm{2}} \right){d}\mathrm{y}={x}\left(\mathrm{5}−\mathrm{3y}\right){dx} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{5}−\mathrm{3y}}{d}\mathrm{y}=\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\Rightarrow\int\frac{\mathrm{1}}{\mathrm{5}−\mathrm{3y}}{d}\mathrm{y}=\int\frac{{x}}{\mathrm{1}+{x}^{\mathrm{2}} }{dx} \\ $$$$\Rightarrow−\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\mid\mathrm{5}−\mathrm{3y}\mid=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)+{k} \\ $$$$\Rightarrow\mathrm{5}−\mathrm{3y}=\frac{{C}}{\:\sqrt{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }}\:\Rightarrow\begin{array}{|c|}{\mathrm{y}=\frac{\mathrm{5}}{\mathrm{3}}−\frac{{C}}{\mathrm{3}\sqrt{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }}}\\\hline\end{array},\:{C}={e}^{{k}} \\ $$

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