form the differential equationfrom the following 1) y=Ae^(3x) +Be^(5x) 2) y^2 =(x−1) 3) c(y+c)^2 +x^3 =0


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Question Number 216787 by Engr_Jidda last updated on 20/Feb/25

$f o r m t h e d i f f e r e n t i a l e q u a t i o n f r o m t h e f o l l o w i n g$ $1) y = {A e}^{3 x} + {B e}^{5 x}$ $2) y^{2} = (x - 1)$ $3) c {(y + c)}^{2} + x^{3} = 0$
	Answered by som(math1967) last updated on 20/Feb/25

	$1 . y = {A e}^{3 x} + {B e}^{5 x}$ $y_{1} = 3 {A e}^{3 x} + 5 {B e}^{5 x}$ $y_{2} = 9 {A e}^{3 x} + 25 {B e}^{5 x}$ $5 y_{1} - y_{2} = 6 {A e}^{3 x}$ $∴ {A e}^{3 x} = \frac{5 y_{1} - y_{2}}{6}$ $a g a i n y_{2} - 3 y_{1} = 10 {B e}^{5 x}$ $∴ {B e}^{5 x} = \frac{y_{2} - 3 y_{1}}{10}$ $\Rightarrow {A e}^{3 x} + {B e}^{5 x} = \frac{5 y_{1} - y_{2}}{6} + \frac{y_{2} - 3 y_{1}}{10}$ $\Rightarrow y = \frac{25 y_{1} - 5 y_{2} + 3 y_{2} - 9 y_{1}}{30} [y = {A e}^{3 x} + {B e}^{5 x}]$ $\Rightarrow y = \frac{16 y_{1} - 2 y_{2}}{30}$ $\Rightarrow y = \frac{8 y_{1} - y_{2}}{15}$ $∴ y_{2} - 8 y_{1} + 15 y = 0$
	Commented by Engr_Jidda last updated on 20/Feb/25

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