solve-the-differential-equation-y-x-p-3-

Question Number 162959 by mkam last updated on 02/Jan/22

s o l v e t h e d i f f e r e n t i a l e q u a t i o n y = x + p^{3}

Commented by mr W last updated on 02/Jan/22

t h i s i s n o t a d i f f e r e n t i a l e q u a t i o n .

Commented by Kamel last updated on 02/Jan/22

M r W p = \frac{d y}{d x} .

Commented by mr W last updated on 02/Jan/22

w h a t i s t h e n p^{3} ?

p^{3} = {(\frac{d y}{d x})}^{3} ? o r p^{3} = y^{‴} = \frac{d^{3} y}{{d x}^{3}} ?

Commented by Kamel last updated on 02/Jan/22

(E) \Leftrightarrow {(\frac{d y}{d x})}^{3} + x = y

Commented by mkam last updated on 02/Jan/22

y e s s i r

Answered by mr W last updated on 02/Jan/22

{(\frac{d y}{d x})}^{3} + x = y

l e t u = y - x

\frac{d u}{d x} = \frac{d y}{d x} - 1

{(\frac{d u}{d x} + 1)}^{3} = u

\frac{d u}{d x} = \sqrt[3]{u} - 1

\frac{d u}{\sqrt[3]{u} - 1} = d x

\int \frac{d u}{\sqrt[3]{u} - 1} = \int d x

l e t t = \sqrt[3]{u} - 1

u = {(t + 1)}^{3}

d u = 3 {(t + 1)}^{2} d t

\int \frac{3 {(t + 1)}^{2}}{t} d t = \int d x

3 \int (t + 2 + \frac{1}{t}) d t = \int d x

3 (\frac{t^{2}}{2} + 2 t + \ln t) + C = x

\frac{{(t + 2)}^{2}}{2} + \ln t + C = \frac{x}{3}

\frac{{(\sqrt[3]{u} + 1)}^{2}}{2} + \ln (\sqrt[3]{u} - 1) + C = \frac{x}{3}

\Rightarrow \frac{{(\sqrt[3]{y - x} + 1)}^{2}}{2} + \ln (\sqrt[3]{y - x} - 1) + C = \frac{x}{3}

Commented by Kamel last updated on 02/Jan/22

T h i s i s a G a u s s d i f f e r e n t i a l e q u a t i o n

f o r e x a m p l e y = 1 + x i s a s i n g u l a r s o l u t i o n .

I f w e c o n s i d e r t h a t p = \frac{d y}{d x} w e g e t a l i n e a r

d i f f e r e n t i a l e q u a t i o n w i t h x, t h e n i f w e s o l v e i t w e g e t

a g e n e r a l s o l u t i o n o f e q u a t i o n w i t h

p a r a m e t e r p = C o n s t .

Commented by mr W last updated on 02/Jan/22

i {d o n}^{'} t u n d e r s t a n d t h a t m u c h . w h a t

i s t h e n t h e s o l u t i o n ? i s a n y t h i n g

w r o n g i n m y s o l u t i o n ?

Commented by Kamel last updated on 02/Jan/22

N o, {i t}^{'} s c o r r e c t .

Commented by mr W last updated on 02/Jan/22

t h a n k s f o r c o n f i r m i n g s i r!

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