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Question Number 128205 by john_santu last updated on 05/Jan/21

$$\mathrm{solve}\frac{dy}{dx}+\frac{dx}{dy}=4x^2+3$$

Answered by mr W last updated on 05/Jan/21

$${\left(y^{\prime }\right)}^2-(4x^2+3)y^{\prime }+1=0$$$$y^{\prime }=\frac{4x^2+3\pm \sqrt{(4x^2+5)(4x^2+1)}}{2}$$$$y=\frac{1}{2}\int [4x^2+3\pm \sqrt{(4x^2+5)(4x^2+1)}]dx$$$$y=\frac{1}{2}[\frac{4x^3}{3}+3x\pm \frac{1}{2}\int \sqrt{({\left(2x\right)}^2+5)({\left(2x\right)}^2+1)}d\left(2x\right)]$$$$withu=2x$$$$y=\frac{2x^3}{3}+\frac{3x}{2}\pm \frac{1}{4}\int \sqrt{(u^2+5)(u^2+1)}du$$$$......$$

Commented by john_santu last updated on 05/Jan/21

Commented by john_santu last updated on 05/Jan/21

$$igotthesameresult$$$$butstuckthesecondintegral$$

Commented by mr W last updated on 05/Jan/21

$$yes,$$$$\int \sqrt{(4x^2+5)(4x^2+1)}dxisnotintegrable$$$$withelementaryfunctions.$$

Commented by mindispower last updated on 05/Jan/21

$$x=\frac{1}{2}sh\left(t\right)$$$$\Rightarrow \int \sqrt{{sh}^2\left(t\right)+5}.\frac{{ch}^2\left(t\right)}{2}dt$$$${ch}^2\left(t\right)=\frac{1+ch\left(2t\right)}{2},\int {ch}^2\left(t\right)=\frac{t}{2}+\frac{1}{4}sh\left(2t\right)$$$$\frac{1}{2}(\frac{t}{2}+\frac{sh\left(2t\right)}{4})\sqrt{{sh}^2\left(t\right)+5}-\frac{1}{4}\int \frac{tsh\left(t\right)ch\left(t\right)}{\sqrt{{sh}^2\left(t\right)+5}}dt-\frac{1}{8}\int \frac{{sh}^2\left(t\right)}{\sqrt{{sh}^2\left(t\right)+5}}dt$$$$\int \frac{tsh\left(t\right)ch\left(t\right)}{\sqrt{{sh}^2\left(t\right)+5}}dt$$$$=\int \frac{tsh\left(2t\right)}{2.\sqrt{4+{ch}^2\left(t\right)}}bypart=\sqrt{4+{ch}^2\left(t\right)}t-\int \sqrt{4+{ch}^2\left(t\right)}dt$$$$\int \sqrt{4+{ch}^2\left(t\right)}dt=\int \sqrt{5+{sh}^2\left(t\right)}dt$$$$t=iw$$$$\Rightarrow \int \sqrt{5+{\left(sh\left(iw\right)\right)}^2}.idw$$$$=i\int \sqrt{5-{sin}^2\left(w\right)}dw$$$$=i\sqrt{5}\int \sqrt{1-{\left(\frac{sinw}{\sqrt{5}}\right)}^2}dw=i\sqrt{5}(E\left(sin(w,\frac{1}{\sqrt{5}})\right)$$$$E(sin\left(\mathrm{\varnothing }\right),k)={\int }_0^{\phi }\sqrt{1-k^2{sin}^2\left(t\right)}dt..2ndelepticintegrak$$$$\int \frac{{sh}^2\left(t\right)}{\sqrt{5+{sh}^2\left(t\right)}},dt=\int \sqrt{{sh}^2\left(t\right)+5}-5\int \frac{dt}{\sqrt{5+{sh}^2\left(t\right)}}$$$$A-5B,AdonneB,t\to ix$$$$B=5i\int \frac{dx}{\sqrt{5-{sin}^2\left(x\right)}}=i\sqrt{5}\int \frac{dx}{\sqrt{1-{\left(\frac{sin\left(x\right)}{\sqrt{5}}\right)}^2}}$$$$=i\sqrt{5}F(sin\left(x\right),\frac{1}{\sqrt{5}})$$$$F(sin\left(t\right),k)=\int \frac{dt}{\sqrt{1-k^{2}sin\left(x\right)}},firstelepticintegral$$$$\Rightarrow \int \sqrt{5+4x^2}.\sqrt{1+4x^2}dxhasclosehardtoexpress$$$$withexvariablebutpossibl$$$$$$$$$$

Commented by john_santu last updated on 05/Jan/21

$$thankyou$$

Commented by mindispower last updated on 06/Jan/21

$$withepleasur$$

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