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Question Number 219243 by mnjuly1970 last updated on 21/Apr/25

Commented by mr W last updated on 23/Apr/25

$$igot\left(solutionseebelow\right)$$$$y=C_1\sin ({\tan }^{-1}x+C_2)$$

Commented by mr W last updated on 22/Apr/25

$$check:$$$$y^{\prime }=C_1\cos ({\tan }^{-1}x+C_2)\times \frac{1}{1+x^2}$$$$\Rightarrow 2x(1+x^2)y^{\prime }=2{xC}_1\cos ({\tan }^{-1}x+C_2)$$$$y^{″}=-C_1\sin ({\tan }^{-1}x+C_2)\times \frac{1}{{(1+x^2)}^2}-C_1\cos ({\tan }^{-1}x+C_2)\frac{2x}{{(1+x^2)}^2}$$$$\Rightarrow {(1+x^2)}^2{y}^{″}=-C_1\sin ({\tan }^{-1}x+C_2)-2{xC}_1\cos ({\tan }^{-1}x+C_2)$$$$\Rightarrow {(1+x^2)}^2{y}^{″}=-y-2x(1+x^2)y^{\prime }$$$$\Rightarrow {(1+x^2)}^2{y}^{″}+2x(1+x^2)y^{\prime }+y=0✓$$

Answered by MathematicalUser2357 last updated on 21/Apr/25

$$y=\frac{x\sqrt{x^2+1}{C}_1+\sqrt{x^2+1}{C}_2}{x^2+1}$$

Commented by mnjuly1970 last updated on 21/Apr/25

$$thxpleasesolutionwithdetails$$

Commented by MathematicalUser2357 last updated on 22/Apr/25

$$\mathrm{solution}\mathrm{with}\mathrm{details}$$$$(\mathrm{so},\mathrm{I}{\mathrm{can}}^{\prime }\mathrm{t}\mathrm{solve}\mathrm{by}\mathrm{hand},\mathrm{so}\mathrm{WolframAlpha}\mathrm{solved}\mathrm{for}\mathrm{me}!)$$

Answered by Ghisom last updated on 21/Apr/25

$$\mathrm{this}\mathrm{one}\mathrm{is}\mathrm{somehow}\mathrm{obvious}[\mathrm{at}\mathrm{least}\mathrm{with}$$$$\mathrm{some}\mathrm{experience}]$$$$\mathrm{multiply}\mathrm{the}\mathrm{equation}\mathrm{with}{(1+x^2)}^{1/2}$$$${(1+x^2)}^{5/2}{y}^{″}+2x{(1+x^2)}^{3/2}{y}^{\prime }+{(1+x^2)}^{1/2}y=0$$$$\mathrm{assume}y=\frac{ax+b}{{(1+x^2)}^{1/2}}$$$$\Rightarrow y^{\prime }=-\frac{bx-a}{{(1+x^2)}^{3/2}}$$$$\Rightarrow y^{″}=\frac{2{bx}^2-3ax-b}{{(1+x^2)}^{5/2}}$$$$\mathrm{inserting}\mathrm{above}\Rightarrow \mathrm{true}$$

Answered by mr W last updated on 22/Apr/25

$$letu=(1+x^2)y^{\prime }$$$$\frac{du}{dx}=(1+x^2)y^{″}+2{xy}^{\prime }$$$$\text{Missing left or extra right}$$$$(1+x^2)\frac{du}{dx}+y=0$$$$(1+x^2)\frac{dy}{dx}\times \frac{du}{dy}+y=0$$$$(1+x^2)y^{\prime }\frac{du}{dy}+y=0$$$$u\frac{du}{dy}+y=0$$$$ydy=-udu$$$$\int ydy=-\int udu$$$$y^2=-u^2+C_1^2$$$$\Rightarrow y^2+u^2=C_1^2$$$$\Rightarrow y=C_1\sin \theta ,u=C_1\cos \theta$$$$y^{\prime }=C_1\cos \theta \frac{d\theta }{dx}$$$$u=(1+x^2)y^{\prime }$$$$C_1\cos \theta =(1+x^2)C_1\cos \theta \frac{d\theta }{dx}$$$$(1+x^2)\frac{d\theta }{dx}=1$$$$d\theta =\frac{dx}{1+x^2}$$$$\int d\theta =\int \frac{dx}{1+x^2}$$$$\theta ={\tan }^{-1}x+C_2$$$$\Rightarrow y=C_1\sin ({\tan }^{-1}x+C_2)✓$$

Commented by mr W last updated on 23/Apr/25

$${\tan }^{-1}x={\sin }^{-1}\frac{x}{\sqrt{1+x^2}}={\cos }^{-1}\frac{1}{\sqrt{1+x^2}}$$$$y=C_1\sin ({\tan }^{-1}x+C_2)$$$$=C_1[\sin \left({\tan }^{-1}x\right)\cos C_2+\cos \left({\tan }^{-1}x\right)\sin C_2]$$$$=C_1[\sin \left({\sin }^{-1}\frac{x}{\sqrt{1+x^2}}\right)\cos C_2+\cos \left({\cos }^{-1}\frac{1}{\sqrt{1+x^2}}\right)\sin C_2]$$$$=C_1[\frac{x}{\sqrt{1+x^2}}\cos C_2+\frac{1}{\sqrt{1+x^2}}\sin C_2]$$$$=\frac{c_{1}x+c_2}{\sqrt{1+x^2}}$$$$thatmeanstheresultcanalso$$$$beexpressedasy=\frac{C_{1}x+C_2}{\sqrt{1+x^2}}.$$

Commented by Ghisom last updated on 23/Apr/25

$$\mathrm{nice}!$$

Commented by mnjuly1970 last updated on 23/Apr/25

$$thanksalotsirW$$

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