I Solved A Non-Homogenous Differential Equation

[@pwmiles56]

From a teaching point of view, it's important that the equation is linear in y. This is what makes it possible to add in multiples of the homogeneous solution to the particular solution.

[@SyberMath]

I agree. I should've said that

[@andreaparma7201]

I like to solve this kind of equations in a slightly different way By differentiating three times, the equation becomes y^(5)-y^(3)=0 (where y^(n) denotes the n-th derivative). This gives y^(3)=h*e^x+k*e^(-x), and by integrating three times y=h*e^x-k*e^(-x)+ax^2+bx+c. We can now substitute in the original equation and get 2a-(ax^2+bx+c)=x^2, which implies (a,b,c)=(-1,0,-2).

[@SyberMath]

Awesome!

[@Drk950]

I did it by the easy way (homogeneous + particular solution), but I liked it the reference to power series at the end

[@MichaelJamesActually]

Idk why, but always loved diff eq. I had also completely forgotten diff eq, so thanks for this!

[@SyberMath]

Np. More is coming...

[@omograbi]

It would be amazing if you devout a special channel for differential equations

[@scottleung9587]

Yeah, I think that would be very cool and helpful since I don't usually encounter problems like these.

[@SyberMath]

Good idea. I'm not that knowledgeable on this topic but could learn along the way (just like @aplusbi)

[@MichaelJamesActually]

It would be cool, but I feel yt already pretty well saturated with differential equations content. I’ve enjoyed the complex numbers channel bc I don’t think those types of problems get much attention.

[@goldfing5898]

I would build the derivative three times: y'' - y = x^2 y''' - y' = 2*x y'''' - y'' = 2 y''''' - y''' = 0 Now we have a homogenous ODE. y''''' = y''' The 5th derivative is equal to the third derivative of the function. Set u = y' v = u' = y'' w = v' = y''' Then we have w'' = y''''' w'' = w Which functions are equal to their second derivative? To my mind come w = k*e^(x) w = k*sinh(x) = k*(e^x - e^(-x))/2 w = k*cosh(x) = k*(e^x + e^(-x))/2 Maybe we should try to generalize this to w(x) = k1*e^x + k2*e^(-x) The natural exponential function for k1 = 1 and k2 = 0. The Sinus hyperbolicus function for k2 = -k1. The casinos hyperbolicus function for k1 = k2. Then we must integrate this w function two times...

[@andirijal9033]

Total Solution = Homogen solution + Partikular Solution

[@giuseppemalaguti435]

Per lomogenea λ^2=1..λ=+1,-1...per la particolare yp=-x^2-2...in sintesi y=c1e^x+c2e^(-x)-(x^2+2)

[@JSSTyger]

My final answer... Yg = Ae^x+Be^(-x)-x²-2 where A and B are unknown constants.

[@kianmath71]

Y = C1e^x + C2e^-x - x^2 - 2😊

[@seanfraser3125]

The homogenous equation is y_h’’ = y_h It’s not difficult to see that y_h = ae^x + be^-x The particular solution is of the form y_p = dx^2 + fx + g Plugging this into the DE, we have 2d - dx^2 - fx - g = x^2 So -d=1, f=0, and 2d-g=0. Thus y_p = 2-x^2 So our general solution is y = ae^x + be^-x + 2-x^2

[@vladimirkaplun5774]

y"+y=x^2 is more interesting. Anyhow without graphs they do not make sense

[@jstone1211]

awful explanation....

[@SyberMath]

why? 😮😄

[@jstone1211]

@@SyberMath within a minute or so you used at least three different variable substitutions. the display used had no space to show previous work. Not a math professor but I did teach electrical engineering courses during my phd....