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Question Number 183532 by MikeH last updated on 26/Dec/22

	Answered by CElcedricjunior last updated on 26/Dec/22
	$3xy′−y=lnx+1 y(1)=−2 (h):3xy′−y=0=>y=e^(∫(1/(3x))dx) =ke^((1/3)ln∣x∣) =ke^((1/3)lnx) pour x∈R_+ ^∗ y=k(x)^(1/3) faisons varier k y(x)=k(x)(x)^(1/3) =>y′(x)=k′(x)(x)^(1/3) +((k(x))/(3(x^2 )^(1/3) )) =>3xy′−y=lnx+1 =>3xy′−y=3xk′(x)(x)^(1/3) +k(x)(x)^(1/3) −k(x)(x)^(1/3) =3xk′(x)(x)^(1/3) <=>k′(x)=((lnx+1)/(3(x^4 )^(1/3) )) <=>k(x)=∫((lnx+1)/(3(x^4 )^(1/3) ))dx=∫((lnx)/(3x(x)^(1/3) ))dx+(1/3)∫x^(−(4/3)) dx =(1/3)((1/(−(4/3)+1))x^(−(4/3)+1) )+(1/3)∫((lnx)/x)×(1/( (x)^(1/3) ))dx posons { ((u=lnx)),((v′=x^(−(4/3)) )) :}=> { ((u′=(1/x))),((v=−(3/( (x)^(1/3) )))) :} =−(1/( (x)^(1/3) ))−((lnx)/( (x)^(1/3) ))+∫x^(−(4/3)) dx =−(4/( (x)^(1/3) ))−((lnx)/( (x)^(1/3) ))+c c∈R....cedric y(x)=k(x)(x)^(1/3) y′(x)=(−(4/( (x)^(1/3) ))−((lnx)/( (x)^(1/3) ))+c)(x)^(1/3) .....junior 3xy′−y=lnx+1 y(1)=−2 (h):3xy′−y=0=>y=e^(∫(1/(3x))dx) =ke^((1/3)ln∣x∣) =ke^((1/3)lnx) pour x∈R_+ ^∗ y=k(x)^(1/3) faisons varier k y(x)=k(x)(x)^(1/3) =>y′(x)=k′(x)(x)^(1/3) +((k(x))/(3(x^2 )^(1/3) )) =>3xy′−y=lnx+1 =>3xy′−y=3xk′(x)(x)^(1/3) +k(x)(x)^(1/3) −k(x)(x)^(1/3) =3xk′(x)(x)^(1/3) <=>k′(x)=((lnx+1)/(3(x^4 )^(1/3) )) <=>k(x)=∫((lnx+1)/(3(x^4 )^(1/3) ))dx=∫((lnx)/(3x(x)^(1/3) ))dx+(1/3)∫x^(−(4/3)) dx =(1/3)((1/(−(4/3)+1))x^(−(4/3)+1) )+(1/3)∫((lnx)/x)×(1/( (x)^(1/3) ))dx posons { ((u=lnx)),((v′=x^(−(4/3)) )) :}=> { ((u′=(1/x))),((v=−(3/( (x)^(1/3) )))) :} =−(1/( (x)^(1/3) ))−((lnx)/( (x)^(1/3) ))+∫x^(−(4/3)) dx =−(4/( (x)^(1/3) ))−((lnx)/( (x)^(1/3) ))+c c∈R....cedric y(x)=k(x)(x)^(1/3) y′(x)=(−(4/( (x)^(1/3) ))−((lnx)/( (x)^(1/3) ))+c)(x)^(1/3) .....junior y(x)=−4−lnx+c(x)^(1/3) y(1)=−2=>−4−ln(1)+c=−2 =>c=2 y(x)=−4−lnx+2((x ))^(1/3) ======================== ..............le celebre cedric junior........ ====================== y(x)=−4−lnx+c(x)^(1/3) y(1)=−2=>−4−ln(1)+c=−2 =>c=2 y(x)=−4−lnx+2((x ))^(1/3) ======================== ..............le celebre cedric junior........ ======================$
	$3 {x y}^{'} - y = l n x + 1 y (1) = - 2$ $(h) : 3 {x y}^{'} - y = 0 => y = e^{\int \frac{1}{3 x} d x} = {k e}^{\frac{1}{3} l n ∣ x ∣} = {k e}^{\frac{1}{3} l n x} p o u r x \in R_{+}^{}$ $y = k \sqrt[3]{x}$ $f a i s o n s v a r i e r k$ $y (x) = k (x) \sqrt[3]{x} => y^{'} (x) = k^{'} (x) \sqrt[3]{x} + \frac{k (x)}{3 \sqrt[3]{x^{2}}}$ $=> 3 {x y}^{'} - y = lnx + 1$ $=> 3 {xy}^{'} - y = 3 {x k}^{'} (x) \sqrt[3]{x} + k (x) \sqrt[3]{x} - k (x) \sqrt[3]{x}$ $= 3 {x k}^{'} (x) \sqrt[3]{x}$ $<=> k^{'} (x) = \frac{l n x + 1}{3 \sqrt[3]{x^{4}}}$ $<=> k (x) = \int \frac{l n x + 1}{3 \sqrt[3]{x^{4}}} d x = \int \frac{l n x}{3 x \sqrt[3]{x}} d x + \frac{1}{3} \int x^{- \frac{4}{3}} d x$ $= \frac{1}{3} (\frac{1}{- \frac{4}{3} + 1} x^{- \frac{4}{3} + 1}) + \frac{1}{3} \int \frac{lnx}{x} \times \frac{1}{\sqrt[3]{x}} dx$ $p o s o n s {\begin{cases} u = l n x \\ v^{'} = x^{- \frac{4}{3}} \end{cases} => {\begin{cases} u^{'} = \frac{1}{x} \\ v = - \frac{3}{\sqrt[3]{x}} \end{cases}$ $= - \frac{1}{\sqrt[3]{x}} - \frac{l n x}{\sqrt[3]{x}} + \int x^{- \frac{4}{3}} d x$ $= - \frac{4}{\sqrt[3]{x}} - \frac{l n x}{\sqrt[3]{x}} + c c \in R . . . . c e d r i c$ $y (x) = k (x) \sqrt[3]{x}$ $y^{'} (x) = (- \frac{4}{\sqrt[3]{x}} - \frac{l n x}{\sqrt[3]{x}} + c) \sqrt[3]{x} . . . . . j u n i o r$ $3 {x y}^{'} - y = l n x + 1 y (1) = - 2$ $(h) : 3 {x y}^{'} - y = 0 => y = e^{\int \frac{1}{3 x} d x} = {k e}^{\frac{1}{3} l n ∣ x ∣} = {k e}^{\frac{1}{3} l n x} p o u r x \in R_{+}^{}$ $y = k \sqrt[3]{x}$ $f a i s o n s v a r i e r k$ $y (x) = k (x) \sqrt[3]{x} => y^{'} (x) = k^{'} (x) \sqrt[3]{x} + \frac{k (x)}{3 \sqrt[3]{x^{2}}}$ $=> 3 {x y}^{'} - y = lnx + 1$ $=> 3 {xy}^{'} - y = 3 {x k}^{'} (x) \sqrt[3]{x} + k (x) \sqrt[3]{x} - k (x) \sqrt[3]{x}$ $= 3 {x k}^{'} (x) \sqrt[3]{x}$ $<=> k^{'} (x) = \frac{l n x + 1}{3 \sqrt[3]{x^{4}}}$ $<=> k (x) = \int \frac{l n x + 1}{3 \sqrt[3]{x^{4}}} d x = \int \frac{l n x}{3 x \sqrt[3]{x}} d x + \frac{1}{3} \int x^{- \frac{4}{3}} d x$ $= \frac{1}{3} (\frac{1}{- \frac{4}{3} + 1} x^{- \frac{4}{3} + 1}) + \frac{1}{3} \int \frac{lnx}{x} \times \frac{1}{\sqrt[3]{x}} dx$ $p o s o n s {\begin{cases} u = l n x \\ v^{'} = x^{- \frac{4}{3}} \end{cases} => {\begin{cases} u^{'} = \frac{1}{x} \\ v = - \frac{3}{\sqrt[3]{x}} \end{cases}$ $= - \frac{1}{\sqrt[3]{x}} - \frac{l n x}{\sqrt[3]{x}} + \int x^{- \frac{4}{3}} d x$ $= - \frac{4}{\sqrt[3]{x}} - \frac{l n x}{\sqrt[3]{x}} + c c \in R . . . . c e d r i c$ $y (x) = k (x) \sqrt[3]{x}$ $y^{'} (x) = (- \frac{4}{\sqrt[3]{x}} - \frac{l n x}{\sqrt[3]{x}} + c) \sqrt[3]{x} . . . . . j u n i o r$ $y (x) = - 4 - l n x + c \sqrt[3]{x}$ $y (1) = - 2 => - 4 - l n (1) + c = - 2$ $=> c = 2$ $y (x) = - 4 - l n x + 2 \sqrt[3]{x}$ $========================$ $. . . . . . . . . . . . . . l e c e l e b r e c e d r i c j u n i o r . . . . . . . .$ $======================$ $y (x) = - 4 - l n x + c \sqrt[3]{x}$ $y (1) = - 2 => - 4 - l n (1) + c = - 2$ $=> c = 2$ $y (x) = - 4 - l n x + 2 \sqrt[3]{x}$ $========================$ $. . . . . . . . . . . . . . l e c e l e b r e c e d r i c j u n i o r . . . . . . . .$ $======================$
	Commented by mr W last updated on 26/Dec/22
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	Commented by MikeH last updated on 26/Dec/22

	$merci$
	Answered by qaz last updated on 27/Dec/22

	$3 {x y}^{'} - y = l n x + 1$ $\Rightarrow y^{'} - \frac{1}{3 x} y = \frac{l n x}{3 x} + \frac{1}{3 x}$ $\Rightarrow y = e^{\int \frac{d x}{3 x}} (C + \int (\frac{l n x}{3 x} + \frac{1}{3 x}) e^{- \int \frac{d x}{3 x}} d x)$ $= \sqrt[3]{x} (C + \int (\frac{l n x}{3 x} + \frac{1}{3 x}) \frac{d x}{\sqrt[3]{x}})$ $= \sqrt[3]{x} (C - \frac{l n x + 4}{\sqrt[3]{x}})$ $= C \sqrt[3]{x} - l n x - 4$ $y (1) = - 2 \Rightarrow C = 2$ $\Rightarrow y = 2 \sqrt[2]{x} - l n x - 4$
	Commented by universe last updated on 27/Dec/22
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