Can you help me solve this initial value problem for nonhomogeneous differential equation?

Question

y" + y' = 2 + 2x + x^2 initial conditions: y(0) = 8, y'(0) = -1

Answer 1

First solve the homogeneous equation

y'' + y' = 0

Here the general solution is y = a*exp(b*x) + c

i.e. y' = a*b*exp(b*x) and y'' = a*b^2*exp(b*x)

So

a*b^2*exp(b*x) + a*b*exp(b*x) = 0

So b + 1 = 0

And b = -1

So y = a*exp(-x) + c - this is not the solution to the equation but has to be added to the solution to obtain the full general solution.


This is not the end of the answer as now we have to consider the nonhomegeneous part

For that the general solution is

y = p*x^3 + q*x^2 + r*x + s

So y' = 3*p*x^2 + 2*q*x + r and y'' = 6*p*x + 2*q

So

x^2*(3*p) + x(2*q + 6*p) + (r+2*q) = x^2 + 2x + 2

So

3*p = 2, so p = 2/3

2*q + 4 = 2, so q = -1

r-2 = 2, so r = 4

So

y = (2/3)*x^3 - x^2 + 4*x + s

And the general solution is

y = a*exp(-x) + (2/3)*x^3 - x^2 + 4*x + c'

c' = c + s

Now as y(0) = 8, so 8 = a + c'

And y'(0) = -1, so -a = -1

so a = 1 and c' = 7

y = exp(-x) + (2/3)*x^3 - x^2 + 4*x + 7

CHECK RESULT:

y' = -exp(-x) + 2*x^2 - 2*x + 4
y" = exp(-x) + 4*x - 2

So y" + y = 2*x^2 + 2*x + 2 - which is correct

Answer 2

I am guessing u havent gone through derivatives yet that why u cant answer this one
here is original question
y" + y' = 2 + 2x + x^2 initial conditions: y(0) = 8, y'(0) = -1

y" is second derivative of y since y'(0)=-1, y"(0)=0 as constasts derivativeis zero always, LHS value is 0+1+8=9 =2 + 2x+x^2
now equation is 9= 2+2x+x^2 so
9-2= 2x+x^2
so 7= 2x+x * x
x^2+2x-7=0
nowI hope u can solve it from here, as 7 wont have factorswhich will give difference of 2, U have to do I dont know whatthat procedure is called in english. i hope it will work let me know what u ccame up with

Answer 3

known answer is complimentary + particular. complimentary section is stumbled on with the help of fixing the linked homogenous utilizing characteristic equation -4m^2 - 4m -a million = 0 ; m1= -a million/2 = m2 ; for repeated roots complimentary = Ae^(-t/2) + Bte^(-t/2) ; utilizing superposition mind-set to undetermined coefficients we come across the kind the certain answer will take with the help of studying 2t^2 -2t + 6e^(-4t) and checking for duplicates interior the complimentary, so our particular kind = Ct^2 + Dt + E + Fe^(-4t) no duplicates, so we determined C,D,E, and F with the help of subing into DE meaning we choose 2d and widely used spinoff of the particular kind (basically particular) y' = 2Ct + D -4Fe^(-4t) ; y''= 2C +16Fe^(-4t) : so then -8C - 64Fe^(-4t) - 8Ct -4D +16Fe^(-4t) - Ct^2 -Dt -E -Fe^(-4t) = 2t^2 -2t +6e^(-4t) ; so -C = 2 :C=-2 : -8C - D = -2 : 16 -D =-2 : -D= -18; D= 18; -49F =6 : F= -6/40 9 ; -8C-4D -E = 0 ; 16 + seventy 2- E = 0 ; E= 88 so the overall answer is y(t) = Ae^(-t/2) +Bte^(-t/2) -2t^2 +18t +88 -(6/40 9)e^(-4t) now we use our initial circumstances meaning we favor to take spinoff so y'(t) = (-A/2)e^(-t/2) +Be^(-t/2) -Bte^(-t/2) -4t +18 +(24/40 9)e^(-4t) ; y(0)=-3=A +88 -(6/40 9) ; A = -4453/40 9 ; y'(0)=0= -A/2 +B +18+(24/40 9) ; B = -895/14 ; hence y(t)= (-3375/40 9)e^(-t/2) - (293/14)te^(-t/2) -2t^2 -14t +seventy 2 -(6/40 9)e^(-4t) ; performed. save in options in case you employ version of parameters you need to divide each little thing with the help of -4 beforehand computing the Wronskian, and also you want not arrive on the same answer to make sure that us both to be superb perfect as there is more beneficial than one answer to any given DE. made a mistake y(0) = -3 and tousled on D so E is erroneous to i imagine i have been given it this time although D= -18 ; E= 88; A= -4453/40 9 ; B= -895/14 ;i checked at that's for confident a answer reason i spent the time to study(guy that took a minute) y(t)= (-4453/40 9)e^(-t/2) -(895/14)te^(-t/2) -2t^2 +18t +88 -(6/40 9)e^(-4t) ; this technique works see you later because the function of t doesnt contain purposes like tan, a million/x, lnx, inverse sin or such issues as that...then you extremely ought to apply version of parameters.

Answer 4

Technology easiest. Other (sucky) options are Laplace transforms, easy to do but labor intensive partial fraction decomposition, and the integrating trick: multiply d.e. by e^x then e^x*y"+e^x*y' = (e^x*y')' so then you integerate to get e^x*y' = integral of RHS but that needs several by parts stages... then divide by e^x and integerate again....uggg

simplify(dsolve({ diff(y(x),x$2)+diff(y(x),x) =2+2*x+x^2, y(0)=8,D(y)(0)= -1},y(x)));
3
y(x) = 2 x + 1/3 x + 5 + 3 exp(-x)

thats 2x+(1/3)*x^3+5+3*e^(-x)

Answer 5

1) solve the homogenous equation, this is standard.
2) find a particular solution by means of the method of the varying constants or the method of similarity
3) fill in the initital conditions, this will give you values for constants that were in your solution by 2)

Answer 6

i've got -x^2=1/2x ...i hope this helps and if it's wrong please tell me the answer.