Key to "Solve the differential equation.y″+(y′)2=2e−y"

nitraiddQ

Answered question

2021-09-18

Solve the differential equation.

y^{″} + {(y^{'})}^{2} = 2 e^{- y}

i1ziZ

Skilled2021-09-19Added 92 answers

Step 1
Given:

y^{″} + {(y^{'})}^{2} = 2 e^{- y}

Step 2
On considering the given differential equation as follows,

y^{″} + {(y^{'})}^{2} = 2 e^{- y}

...(1)
Suppose

u = e^{y}

Then,

y = \log u

y^{'} = \frac{1}{u} \cdot u^{'}

y^{″} = \frac{\cup^{″} - u^{'} \cdot u^{'}}{u^{v}}

Step 3
Equation (1) becomes as follows,

\frac{\cup^{″} - {(u^{'})}^{v}}{u^{v}} + {(\frac{u^{'}}{u})}^{v} = \frac{2}{u}

u''=2
On integrating twice

u^{'} = 2 x + c_{1}

u = x^{2} + 4 x + c_{2}

e^{y} = x^{2} + c_{1} x + c_{2}

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