Recent content by doroulla

i think its something to do with the x on the right hand side but i don't know how to do it. Do i integrate that side? I know when you have a y in the equation you integrate the constants infront of the y. But in this case i don't have this.

hi. I can't figure out this question: d2y/dx2 - 2 dy/dx - 3y = x (i) find complementary function (ii) find particular integral (iii) using (i) and (ii) find the general solution (iv) find the solution that satisfies the initial conditions: y=2/9 at x=0 and dy/dx=-13/3 at x=0 i...

thank you very much. I got it correct now. I found C=-1 so i get: 2x-1 = -e^(2t^2) thus i get : x= 0.5 - 0.5e^(2t^2) when i put t=0 in this equation, i get x=0 thus this is correct right? Thank you

hi. I have asked for help on this again but i got even more confused. The question says solve: dx/dt - 2t(2x-1) = 0 when t=0 when x=0. I did: 0.5 ln(2x-1) = 2t^2 / 2 + K then i have 2x-1 = 0.5(e^t^2) + K putting x=0 and t=0 i get: K=0 thus i have 2x-1 = e^(2t^2) thus...

when t=0 i get x=1 but when i put x=0 i get t=0 this should not make sense right? But if i leave it as k=-2 as i found before as shown above i get x=0 when t=0. So that's is the correct solution?

the question is to find the complementary function, particular integral and general solution of: d2y/dx2 - 2dy/dx - 3y= x i did: m^2 -2m -3 =0 i get (m-3)(m+1)=0 so i have y=Ae^3 + Be^-1 i don't know how to continue from here.. Is this the complementary function?

i forgot about the absolute value. Thank you. Now i get solution: k=0 x= 0.5e2t^2 + 0.5 is this correct? :)

yes i multiplied out. If i put the conditions as it is then i have: ln(-1)=k and this does not exist. You can't have ln(-1), right?

i made a mistake. The C should be Ce^x. Now do you think it makes sense? I didnt understand the other way so i did it the first way toy explained

the question is to solve: dx/dt -2t(2x-1)=0 when t=0, x=0.. I integrated and found: ln(2x-1)=2t^2 + k then : 2x-1 = e^2t^2 -2 then i put in the x=0 when t=0 and i found that k=-2 the solution i found is: x= 0.5e(2t^2) - 0.5 is this correct? Thank you very much

thank you very much! I found general solution is: y=-2x^2 -5x -5 + C is this correct? Or am i completely lost? Thank you for your help

when i was at university i was really good at this. now i can't seem to figure it out. i have been trying for two days so i decided to join a forum. i never thought someone will reply this fast. please do help me out please

what are the correct steps to do? this is wrong way right? but i followed the books

i did that but i get integrating factor = e^-x then i multiplied the original equation with that so i get: (e^-x)dy/dx - (e^-x)y = (e^-x)(x+2x^2) but again i can't integrate this

i want to know the general solution of dy/dx - y = x + 2x^2 i don't know how to do it. looked at every book. i can only do it when seperating the variables but here we have "-y"