Hi ! How do you resolve #y'(t)=ty(t) + 1# and #y(0) =1# without using #erf(x)# please? Thanks!

Question

491 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-09T11:02:44

See below.

#y'(t) = ty(t)+1# is a linear non homogeneous differential equation

For the homogeneous solution we have

#y'_h(t)=t y_h(t)#

multiplying both sides by #y_h(t)# results

#y_h(t)y'_h(t)=t (y_h(t))^2# or

#1/2 d/(dt) (y_h(t))^2= t (y_h(t))^2# or

#(d/(dt)(y_h(t))^2)/ (y_h(t))^2 = 2t#

after integration

#ln(y_h(t))^2 = t^2+C_0# and then

#y_h(t)^2 = C_1 e^(t^2)# and also

#y_h(t) = C_2 e^(t^2/2)#

The particular solution is left as an exercise.

NOTE

Assuming #y_p = C(t)e^(t^2/2)# and substituting into

#y'_p(t) = t y_p(t) + 1# we obtain

#C'(t) = e^(-t^2/2)# and then

#C(t) = int^t e^(- xi^2/2) d xi + C_3# and finally

#y(t) = e^(t^2/2)(int^t e^(- xi^2/2) d xi + C_3)#