Nonlinear Differential Equation

Calculus Level 3

Let us take a look at the following differential equation. The solution of this equation is the function y ( t ) y(t) also depicted as y y

{ y d y d t + y 2 = 10 cos ⁡ ( t ) − 5 sin ⁡ ( t ) y ( 0 ) = 0.5 \large \begin{cases} y \dfrac{dy}{dt} + y^2 = 10\cos(t) - 5\sin(t) \\ y(0) = 0.5 \end{cases}

The closed form solution of this equation is of the following form y ( t ) = ( A cos ⁡ ( t ) − B e − C t D ) 1 E y(t) = \left(A\,\cos (t)-\dfrac{B\,{e}^{-C\,t}}{D}\right)^{\frac{1}{E}} , where, A A , B B , C C , D D , and E E are positive integers with B B and D D being coprime. Enter your answer as: A + B + C + D + E A+B+C+D+E .

Note: I came across a similar equation while framing this problem. I encourage users to attempt this as well.


The answer is 57.

Karan Chatrath by Karan Chatrath , 27, Netherlands

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1 solution

Karan Chatrath
May 26, 2019

Here is one of many ways one can do this problem. However, one step is crucial.

Introducing a change of variables: z = y 2 z = y^2

Implies that: d z d t = 2 y d y d t \frac{dz}{dt} = 2y\frac{dy}{dt}

Replacing this in the differential equation leads us to: 1 2 d z d t + z = 10 cos ⁡ ( t ) − 5 sin ⁡ ( t ) \frac{1}{2}\frac{dz}{dt} + z = 10\cos(t) - 5\sin(t)

So we can see that a nonlinear differential has been converted into a linear differential equation. This is solved using the concept of Laplace transforms. An interested reader can refer to this link.

Remember that the initial condition also gets modified to: z ( 0 ) = 0.25 z(0) = 0.25

Taking the Laplace transform on both sides gives us:

Z ( s ) = 20 s − 10 ( s 2 + 1 ) ( s + 2 ) + 1 4 ( s + 2 ) Z(s) = \frac{20s-10}{(s^2+1)(s+2)} + \frac{1}{4(s+2)}

This expression simplifies to:

Z ( s ) = 10 s ( s 2 + 1 ) − 39 4 ( s + 2 ) Z(s) = \frac{10s}{(s^2+1)} - \frac{39}{4(s+2)}

By referring to any standard table of Laplace transforms, the inverse Laplace transform is computed for the above expression leading to:

z ( t ) = y ( t ) 2 = 10 cos ⁡ ( t ) − 39 4 e − 2 t z(t) = y(t)^2 = 10\cos(t) -\frac{39}{4}e^{-2t}

Therefore:

y ( t ) = ( 10 cos ⁡ ( t ) − 39 4 e − 2 t ) 1 2 \boxed{y(t) = \bigg(10\cos(t) -\frac{39}{4}e^{-2t}\bigg)^{\frac{1}{2}}}

1 Helpful 0 Interesting 2 Brilliant 0 Confused

Great explanation! Several of my steps were "by inspection", so my solution is not very interesting. It's great to see a generalisable tool like the Laplace transform demonstrated here.

Chris Lewis - 2 years ago

Thank you! Glad that you liked the solution. I would still suggest that you post your solution. Sometimes, drawing meaningful conclusions just by inspection can also be very insightful and demonstrate a very deep understanding of the concepts. Knowing multiple approaches to a problem is always helpful.

Karan Chatrath - 2 years ago

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