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Question-77473




Question Number 77473 by BK last updated on 06/Jan/20
Commented by mr W last updated on 06/Jan/20
y=(dy/dx)+(d^2 y/dx^2 )+(d^3 y/dy^3 )+...  (dy/dx)=(d^2 y/dx^2 )+(d^3 y/dy^3 )+...  2(dy/dx)=(dy/dx)+(d^2 y/dx^2 )+(d^3 y/dy^3 )+...  2(dy/dx)=y  ((2dy)/y)=dx  2∫(dy/y)=∫dx  2ln y=x+C_1   ⇒y=Ce^(x/2)
$${y}=\frac{{dy}}{{dx}}+\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\frac{{d}^{\mathrm{3}} {y}}{{dy}^{\mathrm{3}} }+… \\ $$$$\frac{{dy}}{{dx}}=\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\frac{{d}^{\mathrm{3}} {y}}{{dy}^{\mathrm{3}} }+… \\ $$$$\mathrm{2}\frac{{dy}}{{dx}}=\frac{{dy}}{{dx}}+\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }+\frac{{d}^{\mathrm{3}} {y}}{{dy}^{\mathrm{3}} }+… \\ $$$$\mathrm{2}\frac{{dy}}{{dx}}={y} \\ $$$$\frac{\mathrm{2}{dy}}{{y}}={dx} \\ $$$$\mathrm{2}\int\frac{{dy}}{{y}}=\int{dx} \\ $$$$\mathrm{2ln}\:{y}={x}+{C}_{\mathrm{1}} \\ $$$$\Rightarrow{y}={Ce}^{\frac{{x}}{\mathrm{2}}} \\ $$
Commented by BK last updated on 06/Jan/20
thanks
$$\mathrm{thanks} \\ $$

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