How to solve 1st order ODE y'=t+2y, with y(0)=1

Or if u can point me to site with good tutorial. Thanks

Hello,

y' = t + 2y
y' – 2y = t

Solving differential equation without second member:
y' – 2y = 0
y' = 2y
y'/y = 2

Integrating;
ln|y| = 2t + C₀     (with C₀ being the integration constant)
y = e^(2t + C₀)
y = C₁.e^(2t)    (with C₁=e^C₀)

Finding a particular solution of the differential equation:
y' = t + 2y
y' – 2y = t

Setting y=αt+β then y'=α
y' – 2y = α – 2(αt + β) = -2αt + (α – 2β) = t

Thus the system:
{ -2α = 1
{ α – 2β = 0
whose solution is:
{ α = -½
{ β = -¼

Thus the particular solution is:
y = -t/2 – ¼

And the global solution is then the sum of both general and particular solution:
y = C₁.e^(2t) – t/2 – ¼

Solving for initial conditions:
If t=0, y(0)=1
y(0) = C₁.e^(2×0) – 0/2 – ¼ = 1
C₁ × 1 = 1 + ¼
C₁ = 5/4

Thus the final solution:
y(t) = ¼.[5.e^(2t) – 2t – 1]

Check:
y(t) = ¼.[5.e^(2t) – 2t – 1]
y'(t) = ¼.[10.e^(2t) – 2]

y' - 2y = ¼.[10.e^(2t) – 2] – 2×¼.[5.e^(2t) – 2t – 1]
  = ¼.[10.e^(2t) – 2 – 10.e^(2t) + 4t + 2]
  = ¼.[10.e^(2t) – 10.e^(2t) + 4t + 2 – 2]
  = ¼ × 4t
  = t
QED.

Regards,
Dragon.Jade :-)

When solving a DE, you usually want to solve the associated homogenous equation (set all terms without a y equal to 0) first, and then find a specific solution, and add them together.

First, rewrite the equation as y'-2y=t. The homogenous equation in this case is y'-2y=0, or y'=2y. Dividing by y and then integrating by dt gives ln|y|=2t+c, where c is just an integration constant. Take e raised to both sides of the equation to get |y|=ce^(2t) (the c here is a different constant, but it is still a constant). So y=ce^(2t).

For the specific case, notice that t is a polynomial, and derivatives of polynomials give other polynomials (specifically lower order polynomials). Since t is a first order polynomial and we have to add by y, it's safe to assume that y=b+at for some constant a. Plugging this into the original DE gives a-2(b+at)=t. For the equation to be equal for all t, the constant and the co-efficient must be equal on both sides, so a-2b=0 and -2a=1. Solving this system of linear equations gives a=-1/2 and b=-1/4, so -1/4-1/2t is a solution.

Combining the homogenous and specific cases, we get that y=-1/4-1/2t+ce^(2t). Applying the initial condition y(0)=1 gives 1=-1/4+c, so c=5/4. Plugging this in gives y=-1/4-1/2t+5/4e^(2t).

That looks alot like a linear ODE.
in the form of
dy/dt - 2y=t

let me see if i can send you a solution link:
Lets see if that works. Recheck all work.

https://acb0f8a2-a-62cb3a1a-s-sites.goog…




On the very last step, solve for y. sorry. but that is all there is to it. divide by e^-2t at the very end
and you get a soluton for y.



I verified the solution with mathematica. It is right.
Just divide the last equation by e^-2t and you have solve for y.