yy-y-2-e-ax-find-genral-solution-

Question Number 126791 by BHOOPENDRA last updated on 24/Dec/20

{y y}^{″} - {(y^{'})}^{2} = e^{a x} f i n d g e n r a l s o l u t i o n ?

Commented by BHOOPENDRA last updated on 24/Dec/20

h e l p m e o u t t h i s ?

Answered by Olaf last updated on 26/Dec/20

{y y}^{″} - y^{' 2} = e^{a x} (1)

Let y = e^{\frac{1}{2} a x} u (2)

y^{'} = e^{\frac{1}{2} a x} (u^{'} + \frac{1}{2} a u)

y^{″} = e^{\frac{1}{2} a x} (u^{″} + {a u}^{'} + \frac{1}{4} a^{2} u)

(1) : (u^{″} + {a u}^{'} + \frac{1}{4} a^{2} u) u - {(u^{'} + \frac{1}{2} a u)}^{2} = 1

u^{″} u - u^{' 2} = 1 (3)

u^{‴} u + u^{″} u^{'} - 2 u^{'} u^{″} = 0

u^{‴} u - u^{'} u^{″} = 0 (4)

(4) \div {u u}^{″} : \frac{u^{‴}}{u^{″}} - \frac{u^{'}}{u} = 0

\ln ∣ u^{″} ∣ - \ln ∣ u ∣ = C_{1}

u^{″} = C_{2} u

u = α e^{\sqrt{C_{2}} x} + β e^{- \sqrt{C_{2}} x} = α e^{C_{3} x} + β e^{- C_{3} x}

u^{'} = α C_{3} e^{C_{3} x} - β C_{3} e^{- C_{3} x}

u^{″} = α C_{3}^{2} e^{C_{3} x} + β C_{3}^{2} e^{- C_{3} x}

(3) : (α C_{3}^{2} e^{C_{3} x} + β C_{3}^{2} e^{- C_{3} x}) (α e^{C_{3} x} + β e^{- C_{3} x}) - {(α e^{C_{3} x} - β e^{- C_{3} x})}^{2} = 1

α^{2} C_{3}^{2} - α^{2} = 0 (5)

β^{2} C_{3}^{2} - β^{2} = 0 (6)

4 α β C_{3}^{2} = 1 (7)

(5) and (6) : C_{3} = \pm 1

(7) : β = \frac{1}{4 α}

u = α e^{x} + \frac{1}{4 α} e^{- x} or α e^{- x} + \frac{1}{4 α} e^{x}

(2) : y = e^{\frac{1}{4} a x} (α e^{x} + \frac{1}{4 α} e^{- x}) or e^{\frac{1}{4} a x} (α e^{- x} + \frac{1}{4 α} e^{x})

Commented by BHOOPENDRA last updated on 26/Dec/20

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