A Challenging Differential Equation | Can You Solve?

[@henrybarber288]

A trick for simplifying the equation is noticing that on the left hand side we have dy’/dx ÷ dy/dx = dy’/dx · dx/dy which simplifies to dy’/dy using the chain rule. This is equal to e^y, so integrating both sides with respect to y gives us y’ = e^y + c. Of course I would say that your method of getting this result is more elegant and rigorous, but this is the first thing that came to mind for me

[@mickodillon1480]

Good thinking there mate.

[@shamilbabayev8405]

In the end if we denote e^y by t then we will obtain quadratic equitation with t variable so we can find t and therefore e^y will be equal to some expression depending on x and 2 constants , finally y will be: y=ln("the expression found from the quadratic equation").

[@mickodillon1480]

Not gonna lie this was a tough one I thought. Nice solution SyberMath.

[@SyberMath]

Thank you!

[@Hobbitangle]

You're mixing the constants c1 and c2 while intregrating the DE second time. BTW. For better understanding and to use the results easier, the constants in the final expression should be defined by initial condition values, i.e. y0 = y(x0), and y1 = y'(x0)

[@lawrencejelsma8118]

Too bad we don't know f(0) and f'(0) initial conditions or knowing the constants c1 and c2. If we did then Laplace Transforms would have solved this easily for us as: [s^2 - sy(0) - y'(0)]F(s) = [s(1/(s^2 + 1)) - y(0)]U(s) and then an inverse Laplace Transform. In Electrical Engineering we know y(0) and y'(0) type initial conditions as initial t=0 currents and capacitance charges in our system to solve these problems for a given U(s) Laplace Transform input!

[@Chrisoikmath_]

c's are not the same. That's why Wolfarm Alpha gives the result with c1 and c2.

[@scottleung9587]

Painful, indeed - but good info on DE!

[@Blaqjaqshellaq]

y can also be presented as ln(c1) - ln[e^(c1*c2)/e^(c1*x) - c1]

[@lawrencejelsma8118]

Obviously a y=Ax^bx type result. Y' = Abx^bx and Y"=Ab^2x^bx 1/b^2 = 1 or b = +/-1 so Y=ax^-x or Y=ax^x result should be before I watch this video (Cos and Sin functions are expressed in exp() forms).

[@UnlimitedMight-le3db]

If Y is equal to 25 and E is equal to 4 then 2 is my answer in this math problem in a numbering alphabetical order 25+25= 50 multiplied 2 equals 100 which is equal to a numbering sentence

[@yoav613]

Noice

[@user-hp4be3to3z]

Too lousy ~