Question Number 68133 by mr W last updated on 05/Sep/19
![solve y′′=y′y](https://www.tinkutara.com/question/Q68133.png)
$${solve}\:{y}''={y}'{y} \\ $$
Answered by mind is power last updated on 05/Sep/19
![⇒y^′ =(y^2 /2)+s solve this we need value[of f′(0) if f′(0)=a>0 ⇒(y^′ /((y^2 /2)+a))=1⇒(√(2/a))arctan((y/2))=x+s y=2tg(((x+s)/( (√(2/a)))))](https://www.tinkutara.com/question/Q68138.png)
$$\Rightarrow{y}^{'} =\frac{{y}^{\mathrm{2}} }{\mathrm{2}}+{s} \\ $$$${solve}\:{this}\:{we}\:{need}\:{value}\left[{of}\:{f}'\left(\mathrm{0}\right)\right. \\ $$$${if}\:{f}'\left(\mathrm{0}\right)={a}>\mathrm{0} \\ $$$$\Rightarrow\frac{{y}^{'} }{\frac{{y}^{\mathrm{2}} }{\mathrm{2}}+{a}}=\mathrm{1}\Rightarrow\sqrt{\frac{\mathrm{2}}{{a}}}{arctan}\left(\frac{{y}}{\mathrm{2}}\right)={x}+{s} \\ $$$${y}=\mathrm{2}{tg}\left(\frac{{x}+{s}}{\:\sqrt{\frac{\mathrm{2}}{{a}}}}\right) \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 06/Sep/19
![thanks sir!](https://www.tinkutara.com/question/Q68152.png)
$${thanks}\:{sir}! \\ $$
Answered by Kunal12588 last updated on 06/Sep/19
![(d/dx)((dy/dx))=(dy/dx)×y ⇒(dy/dx)=∫ydy ⇒(dy/dx)=(y^2 /2)+c⇒(dy/dx)=((y^2 +2c)/2) ⇒x=2∫(dy/(y^2 +2c)) ⇒x=2×(1/( (√(2c))))×tan^(−1) ((y/( (√(2c)))))+k ⇒x+k=(√(2/c))tan^(−1) ((y/( (√(2c))))) ⇒(y/( (√(2c))))=tan((x+k)(√(c/2))) ⇒y=(√(2c)) tan((x+k)(√(c/2)))](https://www.tinkutara.com/question/Q68179.png)
$$\frac{{d}}{{dx}}\left(\frac{{dy}}{{dx}}\right)=\frac{{dy}}{{dx}}×{y} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\int{ydy} \\ $$$$\Rightarrow\frac{{dy}}{{dx}}=\frac{{y}^{\mathrm{2}} }{\mathrm{2}}+{c}\Rightarrow\frac{{dy}}{{dx}}=\frac{{y}^{\mathrm{2}} +\mathrm{2}{c}}{\mathrm{2}} \\ $$$$\Rightarrow{x}=\mathrm{2}\int\frac{{dy}}{{y}^{\mathrm{2}} +\mathrm{2}{c}} \\ $$$$\Rightarrow{x}=\mathrm{2}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}{c}}}×{tan}^{−\mathrm{1}} \left(\frac{{y}}{\:\sqrt{\mathrm{2}{c}}}\right)+{k} \\ $$$$\Rightarrow{x}+{k}=\sqrt{\frac{\mathrm{2}}{{c}}}{tan}^{−\mathrm{1}} \left(\frac{{y}}{\:\sqrt{\mathrm{2}{c}}}\right) \\ $$$$\Rightarrow\frac{{y}}{\:\sqrt{\mathrm{2}{c}}}={tan}\left(\left({x}+{k}\right)\sqrt{\frac{{c}}{\mathrm{2}}}\right) \\ $$$$\Rightarrow{y}=\sqrt{\mathrm{2}{c}}\:{tan}\left(\left({x}+{k}\right)\sqrt{\frac{{c}}{\mathrm{2}}}\right) \\ $$
Commented by Kunal12588 last updated on 06/Sep/19
![is this correct... this is my first ever encounter of differential equation](https://www.tinkutara.com/question/Q68181.png)
$${is}\:{this}\:{correct}…\:{this}\:{is}\:{my}\:{first}\:{ever}\:{encounter} \\ $$$${of}\:{differential}\:{equation} \\ $$
Commented by mr W last updated on 06/Sep/19
![it′s basically right, but you have assumed that c>0. if c<0, the result is different.](https://www.tinkutara.com/question/Q68182.png)
$${it}'{s}\:{basically}\:{right},\:{but}\:{you}\:{have} \\ $$$${assumed}\:{that}\:{c}>\mathrm{0}.\:{if}\:{c}<\mathrm{0},\:{the} \\ $$$${result}\:{is}\:{different}. \\ $$
Commented by Kunal12588 last updated on 07/Sep/19
![if c=0 x=2∫(dy/y^2 )=2((−1)/y)+k ⇒(1/(k−x))=((y/2)) y=(2/(k−x)) if c<0 x=2∫(dy/(y^2 −2c)) ⇒x=2×(1/(2(√(2c))))log(((y−(√(2c)))/(y+(√(2c)))))+k ⇒(x−k)(√(2c))=log(((y−(√(2c)))/(y+(√(2c))))) what should i do now](https://www.tinkutara.com/question/Q68199.png)
$${if}\:{c}=\mathrm{0} \\ $$$${x}=\mathrm{2}\int\frac{{dy}}{{y}^{\mathrm{2}} }=\mathrm{2}\frac{−\mathrm{1}}{{y}}+{k} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{k}−{x}}=\left(\frac{{y}}{\mathrm{2}}\right) \\ $$$${y}=\frac{\mathrm{2}}{{k}−{x}} \\ $$$${if}\:{c}<\mathrm{0} \\ $$$${x}=\mathrm{2}\int\frac{{dy}}{{y}^{\mathrm{2}} −\mathrm{2}{c}} \\ $$$$\Rightarrow{x}=\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}{c}}}{log}\left(\frac{{y}−\sqrt{\mathrm{2}{c}}}{{y}+\sqrt{\mathrm{2}{c}}}\right)+{k} \\ $$$$\Rightarrow\left({x}−{k}\right)\sqrt{\mathrm{2}{c}}={log}\left(\frac{{y}−\sqrt{\mathrm{2}{c}}}{{y}+\sqrt{\mathrm{2}{c}}}\right) \\ $$$${what}\:{should}\:{i}\:{do}\:{now} \\ $$