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Question Number 155537 by SANOGO last updated on 01/Oct/21

	Answered by amin96 last updated on 01/Oct/21
	$1) y=x^r (dy/dx)=rx^(r−1 ) (d^2 y/dx^2 )= {r^2 −r}x^(r−2) x^2 ×(r^2 −r)x^(r−2) −3x×rx^(r−1) +3×x^r =0 x^r (r^2 −r)−3x^r r+3x^r =0 ⇒ r^2 −r−3r+3=0 r^2 −4r+3=0 Δ=16−12=4>0 { ((r_1 =((4+2)/2)=3)),((r_2 =((4−2)/2)=1)) :} ★y_H =c_1 x^r_1 +c_2 x^r_2 ⇒ y=c_1 x^3 +c_2 x★ −−−−−−−−−−−−−−−−−−− 2) { ((y=x^r )),(((dy/dx)=rx^(r−1) )),(((d^2 y/dx^2 )=(r^2 −r)x^(r−2) )) :} x^2 (r^2 −r)x^(r−2) −3xr∙x^(r−1) r+13x^r =0 x^r (r^2 −r)−3rx^r +13x^r =0 r^2 −r−3r+13=0 ⇒ r^2 −4r+13=0 Δ=16−52=−36<0 { ((r_1 =((4+6i)/2)=2+3i)),((r_2 =2−3i)) :} ⇒ Re{r}=2=a Im{r}=3=b Y=c_1 x^a cos(bln(x))+c_2 x^a sin(bln(x)) ★ Y=c_1 x^2 cos (3ln(x))+c_2 x^2 sin (3ln(x)) Koshi−Euler diferential equation math.Amin$
	$1) y = x^{r} \frac{d y}{d x} = {r x}^{r - 1} \frac{d^{2} y}{{d x}^{2}} = {r^{2} - r} x^{r - 2}$ $x^{2} \times (r^{2} - r) x^{r - 2} - 3 x \times {r x}^{r - 1} + 3 \times x^{r} = 0$ $x^{r} (r^{2} - r) - 3 x^{r} r + 3 x^{r} = 0 \Rightarrow r^{2} - r - 3 r + 3 = 0$ $r^{2} - 4 r + 3 = 0$ $Δ = 16 - 12 = 4 > 0 {\begin{cases} r_{1} = \frac{4 + 2}{2} = 3 \\ r_{2} = \frac{4 - 2}{2} = 1 \end{cases}$ $★ y_{H} = c_{1} x^{r_{1}} + c_{2} x^{r_{2}} \Rightarrow y = c_{1} x^{3} + c_{2} x ★$ $- - - - - - - - - - - - - - - - - - -$ $2) {\begin{cases} y = x^{r} \\ \frac{d y}{d x} = {r x}^{r - 1} \\ \frac{d^{2} y}{{d x}^{2}} = (r^{2} - r) x^{r - 2} \end{cases}$ $x^{2} (r^{2} - r) x^{r - 2} - 3 x r \cdot x^{r - 1} r + 13 x^{r} = 0$ $x^{r} (r^{2} - r) - 3 {r x}^{r} + 13 x^{r} = 0$ $r^{2} - r - 3 r + 13 = 0 \Rightarrow r^{2} - 4 r + 13 = 0$ $Δ = 16 - 52 = - 36 < 0$ ${\begin{cases} r_{1} = \frac{4 + 6 i}{2} = 2 + 3 i \\ r_{2} = 2 - 3 i \end{cases} \Rightarrow R e {r} = 2 = a I m {r} = 3 = b$ $Y = c_{1} x^{a} c o s (b l n (x)) + c_{2} x^{a} s i n (b l n (x))$ $★ Y = c_{1} x^{2} \cos (3 l n (x)) + c_{2} x^{2} \sin (3 l n (x))$ $K o s h i - E u l e r d i f e r e n t i a l e q u a t i o n$ $m a t h . A m i n$
	Commented by SANOGO last updated on 02/Oct/21

	$m e r c i b i e n l e g e n e r a l$
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