Solve differential equation y'-2y= 1-2x

ruigE 2021-01-08 Answered
Solve differential equationy2y=12x
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Szeteib
Answered 2021-01-09 Author has 102 answers

dy/dx+p(x)y=Q(x)
I.F.=ep(x)dx
y(I.F.)=(I.F.)Q(x)+C
dy/dx2y=(12x)
p(x)= -2, Q(x)= (1-2x)
I.F.=ep(x)dx
=e2dx
ye2x=e2x(12x)dx+c
ye2x=(12x)e2x/2(2)e2x/4+c (: using by parts)
ye2x=((2x1)/2)e2x+e2x/2+C
y=((2x1)/2)+1/2+ce2x

Not exactly what you’re looking for?
Ask My Question
Jeffrey Jordon
Answered 2021-12-25 Author has 2087 answers

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2021-05-01
Solve differential equationxy+2y=x3+x, y(1)=2
asked 2022-05-21
g ( x ) is continuous on [ 1 , 2 ] such that g ( 1 ) = 0 and
g ( x ) x 2 = 1 g 2 ( x )
Find g ( 2 )
I found that g ( x ) = sin ( c 1 x ) and since g ( 1 ) = 0 shouldn't my c be equal to 1, so that sin ( 1 1 ) = 0.
But when I try for g ( 2 ) with c = 1, I am getting a different answer from the textbook.
asked 2022-01-22
Find the general solution to these first order differential equations.
3(3x2+у2)dx2xydy=0
asked 2020-12-24
Solve differential equation 2xy9x2+(2y+x2+1)dydx=0, y(0)=3
asked 2022-01-22
Important
Solve the bernoullis
asked 2022-06-22
Solve the differential equation by the appropriate substitution
x 2 d y d x + 2 x y = x 4 y 2 + 1
asked 2022-02-16
I've tried many times to reach the solution of a first order differential equation (of the last equation) but unfortunately I couldn't. Could you please help me to know how did he get this solution.
The equation is:
dXdt=kf(kf+kb)X
The author assumed that
X=0 at t=0
Then, he said in his first publication the solution is given as:
X=kf(kf+kb){1exp[(kf+kb)t]}
I agree with him about the above equation as I am able to reach this form by using the normal solution of the first order differential equation.
In another publication, the same author dealt with the same equations, but when he wrote the solution of the first order differential equation, he wrote it in another form which I could't reach it and I don't know how did he reach it. This form is given as:
X=kf(kf+kb){1exp[(kf+kb)t]}+X0 exp[(kf+kb)t] and he said X0 was taken to be zero in the first publication.
Can any of you help how did the author get this solution please?