Non-Linear Differential Equation

Calculus Level 4

1 4 ( y ′ ) 2 − 1 2 y ′ = y 2 − y \large \frac{1}{4}(y')^2 - \frac{1}{2}y' = y^2 - y

The non-linear differential equation above admits two non-zero solutions y 1 y_1 and y 2 y_2 . Suppose that y 1 ( 0 ) = y 2 ( 0 ) = 2 y_1(0) = y_2(0) = 2 .

What is y 1 ( 1 ) + y 2 ( 1 ) y_1(1) + y_2(1) ?


The answer is 15.913.

Andrew Ellinor by Andrew Ellinor , 33, USA

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3 solutions

Michael Mendrin
Nov 9, 2015

This quickly factors as follows, into two easily solved first order differential equations

1 4 ( y ′ ) 2 − y 2 = ( 1 2 y ′ + y ) ( 1 2 y ′ − y ) = 1 2 y ′ − y \dfrac { 1 }{ 4 } { \left( y' \right) }^{ 2 }-{ y }^{ 2 }=\left( \dfrac { 1 }{ 2 } y'+y \right) \left( \dfrac { 1 }{ 2 } y'-y \right) =\dfrac { 1 }{ 2 } y'-y

The art of solving differential equations is still too much like alchemy, too much depends on luck

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Thanks for bringing up a discussion @Michael Mendrin . I'll pick up where you left off!

Case I: 1 2 y ′ − y = 0 \dfrac { 1 }{ 2 } y'-y = 0 . In this case, we can solve using separation of variable methods . 1 2 y ′ − y = 0 ⟶ d y y = 2 d x ⟶ ln ⁡ ( y ) = 2 x + C ⟶ y = C e 2 x \frac { 1 }{ 2 } y'-y = 0 \longrightarrow \frac{dy}{y} = 2dx \longrightarrow \ln(y) = 2x + C \longrightarrow y = Ce^{2x} In this instance, y ( 0 ) = 2 y(0) = 2 implies that our first non-trivial solution is y 1 = 2 e 2 x y_1 = 2e^{2x} .

Case II: 1 2 y ′ − y ≠ 0 \dfrac { 1 }{ 2 } y'-y \neq 0 . In this case, we safely divide through by 1 2 y ′ − y \dfrac { 1 }{ 2 } y'-y , which returns the equation 1 2 y ′ + y = 1 \dfrac { 1 }{ 2 } y'+y = 1 . Using the method of separation of variables again we can easily find y 2 y_2 :

1 2 y ′ + y = 1 ⟶ d y 1 − y = 2 d x ⟶ − ln ⁡ ∣ 1 − y ∣ = 2 x + C ⟶ y = C e − 2 x + 1 \frac { 1 }{ 2 } y'+y = 1 \longrightarrow \frac{dy}{1 - y} = 2dx \longrightarrow -\ln\left|1 - y\right| = 2x + C \longrightarrow y = Ce^{-2x} + 1

Here, y ( 0 ) = 2 y(0) = 2 implies that our second non-trivial solution is y 2 = e − 2 x + 1 y_2 = e^{-2x} + 1 . So our final answer is

y 1 ( 1 ) + y 2 ( 1 ) = 2 e 2 + e − 2 + 1 ≈ 15.913 y_1(1) + y_2(1) = 2e^2 + e^{-2} + 1 \approx \boxed{15.913}

Andrew Ellinor - 5 years, 7 months ago

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What I can say is that if, for example, that 1 4 \dfrac {1}{4} was 1 3 \dfrac {1}{3} instead, this thing would just blow up. I'd hate to try to imagine the full solution.

Michael Mendrin - 5 years, 7 months ago
Tom Engelsman
Mar 4, 2017

Let u = y ′ u = y' and multiply the DE through by 4 to obtain:

u 2 − 2 u − 4 ( y 2 − y ) = 0 ⇒ u = 2 ± 4 + 16 ( y 2 − y ) 2 = 2 ± 2 4 y 2 − 4 y + 1 2 = 1 ± ( 2 y − 1 ) 2 = 1 ± ( 2 y − 1 ) u^2 - 2u - 4(y^2 - y) = 0 \Rightarrow u = \frac{2 \pm \sqrt{4 + 16(y^2 - y)}}{2} = \frac{2 \pm 2\sqrt{4y^2 - 4y + 1}}{2} = 1 \pm \sqrt{(2y-1)^2} = 1 \pm (2y - 1)

or y ′ = 2 y , 2 − 2 y y' = 2y, 2 - 2y . Integrating both DE's now produces:

l n ∣ y 1 ( x ) ∣ = 2 x + C 1 ; y 1 ( 0 ) = 2 ⇒ C 1 = l n ( 2 ) ⇒ y 1 ( x ) = 2 e 2 x ln|y_1(x)| = 2x + C_1; y_1(0) = 2 \Rightarrow C_1 = ln(2) \Rightarrow y_1(x) = 2e^{2x}

− l n ∣ 1 − y 2 ( x ) ∣ = 2 x + C 2 ; y 2 ( 0 ) = 2 ⇒ C 2 = 0 ⇒ y 2 ( x ) = 1 + e − 2 x -ln|1-y_2(x)| = 2x + C_2; y_2(0) = 2 \Rightarrow C_2 = 0 \Rightarrow y_2(x) = 1 + e^{-2x}

Finally, y 1 ( 1 ) + y 2 ( 1 ) = 2 e 2 + 1 + e − 2 ≈ 15.913 . y_1(1) + y_2(1) = 2e^2 + 1 + e^{-2} \approx \boxed{15.913}.

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You missed a third solution the constant function y=1

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