Question Number 102716 by Ar Brandon last updated on 10/Jul/20

y′′+3y′−10y=14e^(−5x)

Answered by bramlex last updated on 10/Jul/20

Homogenous   r^2 +3r−10=0 ⇒(r+5)(r−2)=0  y_h  = Ae^(2x) +Be^(−5x)   PI. y_p  = Cxe^(−5x)  ; y′ = Ce^(−5x) −5Cxe^(−5x)   y′′=−5Ce^(−5x) −5C(e^(−5x) −5xe^(−5x) )  y′′=−10Ce^(−5x) +25Cxe^(−5x)   comparing coefficient  −10Ce^(−5x) +25Cxe^(−5x) +3Ce^(−5x) −15Cxe^(−5x) −10Cxe^(−5x) =14e^(−5x)   ⇒−7Ce^(−5x) =14e^(−5x)  ⇒C=−2  y_p =−2xe^(−5x)   ∴generall solution   y= Ae^(2x) +Be^(−5x) −2xe^(−5x)    ▼^(°⌢•⌢)

Commented byAr Brandon last updated on 10/Jul/20

πŸ˜… Thanks

Commented bybramlex last updated on 10/Jul/20

because e^(−5x)  is homogenous  solution

Commented byAr Brandon last updated on 10/Jul/20

I spent my time trying to have the solution using  y_p =λe^(−5x)   But why is it instead y_p =Cxe^(−5x)  ?

Commented byAr Brandon last updated on 10/Jul/20

And... how does that affect your choice of y_p ?

Commented byAr Brandon last updated on 10/Jul/20

OK I think I′ve understood. I completely forgot   about that. The form λe^(−5x)  is already present in  the Complementary function and it′s pointless having  it again in y_p . So we therefore move to the higher  degree of x.

Commented byAr Brandon last updated on 10/Jul/20

😁 Thanks

Commented byAr Brandon last updated on 10/Jul/20

https://www.pdfdrive.com/search?q=jee+mathematics&pagecount=&pubyear=&searchin=

Commented byDwaipayan Shikari last updated on 10/Jul/20

Do you study Indian I IT  JEE math book sir?

Commented byAr Brandon last updated on 10/Jul/20

Yes bro,😁 They're among my ebooks.πŸ˜ƒ

Commented byDwaipayan Shikari last updated on 10/Jul/20

Oh I am a Indian student alsoπŸ˜ƒπŸ˜›.Which book do you study?😁

Commented byAr Brandon last updated on 10/Jul/20

😁From your name I could deduce that.

Commented byAr Brandon last updated on 10/Jul/20

But actually I just use these JEE Ebooks to help maximise my skills. I noticed it contains questions which push you to the peak of reasoning in every chapter.And I'm also a student.πŸ˜€

Commented byDwaipayan Shikari last updated on 10/Jul/20

πŸ‘½ I am also.......( understand what I want to mean)

Commented byAr Brandon last updated on 10/Jul/20

πŸ‘πŸ˜

Commented byIRAN_majid last updated on 10/Jul/20

ok

Answered by mathmax by abdo last updated on 10/Jul/20

wronskien method  he→y^(′′) +3y^′ −10y =0 ⇒r^2  +3r−10 =0  Δ =9−4(−10) =49 ⇒r_1 =((−3+7)/2) =2 and r_2 =((−3−7)/2) =−5 ⇒  ⇒y_h =a e^(2x)  +b e^(−5x)  =au_1  +bu_2   w(u_1 ,u_2 ) = determinant (((e^(2x)             e^(−5x) )),((2e^(2x)          −5 e^(−5x) ))) =−5 e^(−3x) −2e^(−3x)  =−7 e^(−3x)   W_1 = determinant (((0                          e^(−5x) )),((14 e^(−5x)            −5e^(−5x) ))) =−14 e^(−10x)   W_2 =  determinant (((e^(2x)               0)),((2e^(2x)        14 e^(−5x) ))) =14e^(−3x)   v_1 =∫ (w_1 /W)dx =∫  ((−14 e^(−10x) )/(−7 e^(−3x) )) dx = 2 ∫ e^(−7x)  dx =−(2/7)e^(−7x)   v_2 =∫ (W_2 /W) dx =∫ ((14e^(−3x) )/(−7e^(−3x) ))dx =−2 x ⇒ y_p =u_1 v_1  +u_2 v_2   =e^(2x) ×(−(2/7))e^(−7x)  + e^(−5x) ×(−2x) =−(2/7) e^(−7x)  −2x e^(−5x)  ⇒  the general solution is y =y_h  +y_p   y = ae^(2x)  +b e^(−5x)  −(2/7) e^(−7x)  −2x e^(−5x)

Commented byAr Brandon last updated on 10/Jul/20

What's the theory, Sir ? πŸ˜ƒ

Commented byAr Brandon last updated on 10/Jul/20

I understand you may be so busy by now. Please reply whenever you feel it's OK to do so.πŸ˜ƒ

Commented bymathmax by abdo last updated on 10/Jul/20

wronskien method

Answered by Aziztisffola last updated on 10/Jul/20