The General Solution of the DE y''+y'=tan(t) ?

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SUMMARY

The general solution to the differential equation y'' + y' = tan(t) involves first solving the associated homogeneous equation, yielding solutions y1 = 1 and y2 = e^(-t). The general solution is expressed as y = u1*y1 + u2*y2, where u1 = -ln|cos(t)| + c1 and u2 is derived from the integral -∫e^t*tan(t) dt. This integral does not have a closed form, leading to the suggestion of defining a new function, Λ(t) = ∫_0^t tan(s) e^s ds, to represent the solution.

PREREQUISITES
  • Understanding of differential equations, specifically second-order linear equations.
  • Familiarity with the method of variation of parameters.
  • Knowledge of the Wronskian and its application in solving differential equations.
  • Basic calculus, particularly integration techniques involving trigonometric functions.
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  • Learn about the Wronskian and its significance in determining the linear independence of solutions.
  • Explore the concept of expressing solutions in terms of special functions when closed forms are not available.
  • Investigate series solutions for differential equations, particularly for integrals that cannot be solved analytically.
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Ahmad Obeid
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Hello There, I hope I posted this in the right thread.
I've been struggling with solving this particular Differential Equation and just couldn't find any way to solve it completely..

1. Homework Statement

I am only required to find the general solution of the differential equation
y'' + y' = tan(t)

Homework Equations


Wronskian of two functions.
Characteristic Polynomial of a (homogeneous) Differential Equation.

The Attempt at a Solution


First I found the solution to the associated DE y''+y'=0 which gave me y1=1 and y2=e-t
Thus the general solution is y= u1*y1 + u2*y2 where u1 and u2 are two functions of t to be determined.
Using the method of variation of parameters I ended up with u1 = -ln|cos(t)| + c1 (Note that the Wronskian of y1 & y2 is -e-t )
However I ended up with u2 = -∫et*tant dt which is obviously unsolvable..
You can find attached my work and attempts.
Is there anything wrong? or is there another way around? like just writing the integral as an infinite series?
Thank you for your help!
 

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Ahmad Obeid said:
Hello There, I hope I posted this in the right thread.
I've been struggling with solving this particular Differential Equation and just couldn't find any way to solve it completely..

1. Homework Statement

I am only required to find the general solution of the differential equation
y'' + y' = tan(t)

Homework Equations


Wronskian of two functions.
Characteristic Polynomial of a (homogeneous) Differential Equation.

The Attempt at a Solution


First I found the solution to the associated DE y''+y'=0 which gave me y1=1 and y2=e-t
Thus the general solution is y= u1*y1 + u2*y2 where u1 and u2 are two functions of t to be determined.
Using the method of variation of parameters I ended up with u1 = -ln|cos(t)| + c1 (Note that the Wronskian of y1 & y2 is -e-t )
However I ended up with u2 = -∫et*tant dt which is obviously unsolvable..
You can find attached my work and attempts.
Is there anything wrong? or is there another way around? like just writing the integral as an infinite series?
Thank you for your help!

Lots of problems have solutions that cannot be written in "closed form"; perhaps this is one of them. What I mean is that you can invent a new function ##\Lambda(t) = \int_0^t \tan(s) e^s \, ds## and can then express your answer in terms of ##\Lambda(\cdot)##.
 
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Likes   Reactions: Ahmad Obeid
Oh so I just leave it as is? Thought there was some extra step I should make...
If only that y' was a y life would've been much easier haha
Thank you for your help !
 

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