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Question: Does anybody know how to solve this?...\(yy''={{y}^{2}}y'+y{{'}^{2}}\)...

Does anybody know how to solve this?...yy′′=y2y′+y′2yy''={{y}^{2}}y'+y{{'}^{2}}

Explanation

Solution

From the question we have been asked to solve the yy′′=y2y′+y′2yy''={{y}^{2}}y'+y{{'}^{2}}. To solve this question we will use the differentiation and integration. SO, first we will use the trick ⇒y′′=d2ydx2=dy′dy×dydx=dy′dy×y′\Rightarrow y''=\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{dy'}{dy}\times \dfrac{dy}{dx}=\dfrac{dy'}{dy}\times y' and simplify the equation. After doing it we will use integration form that variable and separable form that is by finding the integration factor and doing the integration. Thus we will proceed our solution as follows.

Complete step-by-step answer:
We know that,
⇒yy′′=y2y′+y′2\Rightarrow yy''={{y}^{2}}y'+y{{'}^{2}}
Here we can say that yy is a constant and it is obviously a trivial solution as every term has a derivative in it.
Now, we will modify the formula using a common physics trick.
⇒y′′=d2ydx2=dy′dy×dydx=dy′dy×y′\Rightarrow y''=\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{dy'}{dy}\times \dfrac{dy}{dx}=\dfrac{dy'}{dy}\times y'
By substituting this we get equation reduced as follows.
⇒ydy′dyy′=y2y′+y′2\Rightarrow y\dfrac{dy'}{dy}y'={{y}^{2}}y'+y{{'}^{2}}
⇒dy′dy−y′y=y\Rightarrow \dfrac{dy'}{dy}-\dfrac{y'}{y}=y
Now, we will do integration. First we will find the integrating factor I(y)I\left( y \right):
The integrating factor will be I.F=e∫p(x)dxI.F={{e}^{\int{p\left( x \right)dx}}} for a equation of form dydx+p(x)y=q(x)\dfrac{dy}{dx}+p\left( x \right)y=q\left( x \right).
I.F=e∫p(x)dxI.F={{e}^{\int{p\left( x \right)dx}}}
⇒I(y)=e∫−1ydy\Rightarrow I\left( y \right)={{e}^{\int{-\dfrac{1}{y}dy}}}
⇒I(y)=e(∫−ln⁡y+ln⁡c)=cy\Rightarrow I\left( y \right)={{e}^{\left( \int{-\ln y+\ln c} \right)}}=\dfrac{c}{y}
Now, we will do integration to both sides of equation.
⇒I(dy′dy−y′y=y)\Rightarrow I\left( \dfrac{dy'}{dy}-\dfrac{y'}{y}=y \right)
⇒1ydy′dy−y′y2=1\Rightarrow \dfrac{1}{y}\dfrac{dy'}{dy}-\dfrac{y'}{{{y}^{2}}}=1
⇒∫(1ydy′dy−y′y2)=∫1\Rightarrow \int{\left( \dfrac{1}{y}\dfrac{dy'}{dy}-\dfrac{y'}{{{y}^{2}}} \right)}=\int{1}
Here we integrate the above equation using the formula ∫1=y+c\int{1}=y+c.
⇒(1yy′)′=1\Rightarrow \left( \dfrac{1}{y}y' \right)'=1
⇒1yy′=y+c\Rightarrow \dfrac{1}{y}y'=y+c
⇒y′=y(y+c)\Rightarrow y'=y\left( y+c \right)
This is separable. So, we writhe above equation as follows.
⇒dydx=y(y+c)\Rightarrow \dfrac{dy}{dx}=y\left( y+c \right)
⇒dyy2+cy=dx\Rightarrow \dfrac{dy}{{{y}^{2}}+cy}=dx
⇒1cln⁡(yy+c)=x+d\Rightarrow \dfrac{1}{c}\ln \left( \dfrac{y}{y+c} \right)=x+d
⇒yy+c=ec(x+d)\Rightarrow \dfrac{y}{y+c}={{e}^{c\left( x+d \right)}}
⇒y(1−ec(x+d))=cec(x+d)\Rightarrow y\left( 1-{{e}^{c\left( x+d \right)}} \right)=c{{e}^{c\left( x+d \right)}}
⇒y=cec(x+d)(1−ec(x+d))\Rightarrow y=\dfrac{c{{e}^{c\left( x+d \right)}}}{\left( 1-{{e}^{c\left( x+d \right)}} \right)}
Therefore, the solution will be ⇒y=cec(x+d)(1−ec(x+d))\Rightarrow y=\dfrac{c{{e}^{c\left( x+d \right)}}}{\left( 1-{{e}^{c\left( x+d \right)}} \right)}.

Note: Students must be very careful in calculations. Students must have good knowledge in the concept of differentiation and integration. We must not do mistake in finding the integration factor wrong. For example the integrating factor for any equation reduced to form dydx+p(x)y=q(x)\dfrac{dy}{dx}+p\left( x \right)y=q\left( x \right) will be I.F=e∫p(x)dxI.F={{e}^{\int{p\left( x \right)dx}}}. So, for this question if we will substitute ⇒I(y)=e∫−ydy\Rightarrow I\left( y \right)={{e}^{\int{-ydy}}} this instead of this ⇒I(y)=e∫−1ydy\Rightarrow I\left( y \right)={{e}^{\int{-\dfrac{1}{y}dy}}} our whole solution will be wrong.