A not so easy differential equation

Calculus Level 5

d y d x ⋅ d 3 y d x 3 = 3 ( d 2 y d x 2 ) 2 \dfrac { dy }{ dx } \cdot \dfrac { { d }^{ 3 }y }{ d{ x }^{ 3 } } =3{ \left( \dfrac { { d }^{ 2 }y }{ d{ x }^{ 2 } } \right) }^{ 2 } The general solution of the differential equation above is of the form y α = k x + f β g + c { y }^{ \alpha }=\dfrac { \sqrt [ \beta ]{ kx+f } }{ g } +c , where k , k, f , f, g , g, and c c are arbitrary constants and α \alpha and β \beta are coprime positive integers .

Find the value of α + β . \ \alpha +\beta.


The answer is 3.

Rishi Sharma by Rishi Sharma , 32, India

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1 solution

Rishi Sharma
Jun 30, 2016

Let d y d x = p \frac{ dy }{ dx } = p . Then the equation transforms to p d 2 p d x 2 = 3 ( d p d x ) 2 p\frac { { d }^{ 2 }p }{ d{ x }^{ 2 } } =3{ \left( \frac { dp }{ dx } \right) }^{ 2 } Therefore d 2 p d x 2 d p d x = 3 p d p d x ⇒ d d x { ln ⁡ d p d x } = 3 p d p d x \displaystyle \frac { \frac { { d }^{ 2 }p }{ d{ x }^{ 2 } } }{ \frac { dp }{ dx } } =\frac { 3 }{ p } \frac { dp }{ dx } \Rightarrow \frac { d }{ dx } \left\{ \ln { \frac { dp }{ dx } } \right\} =\frac { 3 }{ p } \frac { dp }{ dx } .Now integrating both sides, we get ln ⁡ d p d x = 3 ln ⁡ p + ln ⁡ a ⇒ d p d x = a p 3 \ln { \frac { dp }{ dx } } =3\ln { p } +\ln { a } \Rightarrow \frac { dp }{ dx } =a{ p }^{ 3 } Again integrating both the sides and substituting p = d y d x p = \frac { dy }{ dx } we get, d y d x = − 1 2 ( a x + b ) ⇒ y = − 2 a x − b − a + c \frac { dy }{ dx } =\sqrt { \frac { -1 }{ 2\left( ax+b \right) } } \Rightarrow \quad y\quad =\quad \frac { \sqrt { -2ax-b } }{ -a } +c (a,b,c are arbitrary constants.) Hence α = 1 \alpha = 1 and β = 2 \beta = 2 . So α + β = 1 + 2 = 3 \alpha + \beta = 1+2 = \boxed { 3 }

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Moderator note:

Great approach. The initial substitution of p p simplifies the approach greatly.

There should be one more general solution. From your first equation, you divide by dp/dx on both sides. But dp/dx = 0 could be a solution, which leads to p = C and y = Cx+D, where C and D are constants. It's easy to check that it satisfies the original differential equation.

Another minor point, on your general solution, constant g could be absorbed into constant k and f, making it only 3 independent constants, which makes sense for a third order differential equation.

Wei Chen - 4 years, 11 months ago

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Thanks didn't noticed them first.

Rishi Sharma - 4 years, 10 months ago

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