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Question Number 1575 by 112358 last updated on 21/Aug/15

Solve the following D.E. ((dy/dx))^2 +2y+2=0 Does ((d^2 y/dx^2 ))^2 +2y+2=0 have any solutions other than y=−1 ?

$${Solve}\:{the}\:{following}\:{D}.{E}. \\ $$$$\:\:\:\:\:\:\:\:\:\:\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} +\mathrm{2}{y}+\mathrm{2}=\mathrm{0}\: \\ $$$${Does}\:\:\left(\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} }\right)^{\mathrm{2}} +\mathrm{2}{y}+\mathrm{2}=\mathrm{0}\:{have} \\ $$$${any}\:{solutions}\:{other}\:{than} \\ $$$${y}=−\mathrm{1}\:? \\ $$

Commented by 123456 last updated on 21/Aug/15

(d/dx)(y(dy/dx))=((dy/dx))^2 +y(d^2 y/dx^2 )

$$\frac{{d}}{{dx}}\left({y}\frac{{dy}}{{dx}}\right)=\left(\frac{{dy}}{{dx}}\right)^{\mathrm{2}} +{y}\frac{{d}^{\mathrm{2}} {y}}{{dx}^{\mathrm{2}} } \\ $$

Answered by prakash jain last updated on 21/Aug/15

y=c_1 x^2 +c_2 x+c_3 y^′ =2c_1 x+c_2 (2c_1 x+c_2 )^2 +2(c_1 x^2 +c_2 x+c_3 )+2=0 4c_1 ^2 x^2 +4c_1 c_2 x+c_2 ^2 +2c_1 x^2 +2c_2 x+2c_3 +2=0 c_1 =−(1/2) or c_1 =0 4c_1 c_2 +2c_2 =0 c_2 ^2 +2c_3 +2=0⇒c_3 =−(c_2 ^2 /2)−1 y(x)=−(1/2)x^2 +c_2 x−(((c_2 ^2 +2)/2)) The second differential equation should result in a general solution with 2 arbitrary constants.

$${y}={c}_{\mathrm{1}} {x}^{\mathrm{2}} +{c}_{\mathrm{2}} {x}+{c}_{\mathrm{3}} \\ $$$${y}^{'} =\mathrm{2}{c}_{\mathrm{1}} {x}+{c}_{\mathrm{2}} \\ $$$$\left(\mathrm{2}{c}_{\mathrm{1}} {x}+{c}_{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{2}\left({c}_{\mathrm{1}} {x}^{\mathrm{2}} +{c}_{\mathrm{2}} {x}+{c}_{\mathrm{3}} \right)+\mathrm{2}=\mathrm{0} \\ $$$$\mathrm{4}{c}_{\mathrm{1}} ^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{4}{c}_{\mathrm{1}} {c}_{\mathrm{2}} {x}+{c}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{2}{c}_{\mathrm{1}} {x}^{\mathrm{2}} +\mathrm{2}{c}_{\mathrm{2}} {x}+\mathrm{2}{c}_{\mathrm{3}} +\mathrm{2}=\mathrm{0} \\ $$$${c}_{\mathrm{1}} =−\frac{\mathrm{1}}{\mathrm{2}}\:{or}\:{c}_{\mathrm{1}} =\mathrm{0} \\ $$$$\mathrm{4}{c}_{\mathrm{1}} {c}_{\mathrm{2}} +\mathrm{2}{c}_{\mathrm{2}} =\mathrm{0} \\ $$$${c}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{2}{c}_{\mathrm{3}} +\mathrm{2}=\mathrm{0}\Rightarrow{c}_{\mathrm{3}} =−\frac{{c}_{\mathrm{2}} ^{\mathrm{2}} }{\mathrm{2}}−\mathrm{1} \\ $$$${y}\left({x}\right)=−\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} +{c}_{\mathrm{2}} {x}−\left(\frac{{c}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{2}}{\mathrm{2}}\right) \\ $$$$\mathrm{The}\:\mathrm{second}\:\mathrm{differential}\:\mathrm{equation}\:\mathrm{should}\:\mathrm{result} \\ $$$$\mathrm{in}\:\mathrm{a}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{with}\:\mathrm{2}\:\mathrm{arbitrary}\:\mathrm{constants}. \\ $$

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