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Question Number 96772 by abdomathmax last updated on 04/Jun/20

solve y^(′′) −2y =x^2 sinx  and y(0)=0 ,y^′ (0) =1

solvey2y=x2sinxandy(0)=0,y(0)=1

Answered by mathmax by abdo last updated on 04/Jun/20

(he) →y^(′′) −2y =0 →r^2 −2 =0 ⇒r =+^− (√2) ⇒y_h =a e^((√2)x)  +b e^(−(√2)x)  =au_1  +bu_2   W(u_1  ,u_2 ) = determinant (((e^(x(√2))               e^(−x(√2)) )),(((√2)e^(x(√2))        −(√2)e^(−(√2)x) )))=−2(√2)  W_1 = determinant (((o          e^(−x(√2)) )),((x^2 sinx      −(√2)e^(−(√2)x) )))=−x^2 sinx e^(−x(√2))   W_2 = determinant (((e^(x(√2))            0)),(((√2)e^(x(√2))      x^2 sinx)))=x^2 sinx e^(x(√2))   v_1 =∫  (W_1 /W)dx  =(1/(2(√2)))∫  x^2  sinx e^(−x(√2)) dx =(1/(2(√2))) Im(∫ x^2  e^(ix−x(√2)) dx)  ∫ x^2  e^((i−(√2))x)  dx =_(byparts)     (1/(i−(√2))) x^2  e^((i−(√2))x)  −(1/(i−(√2)))∫ 2x e^((i−(√2))x) [dx  =(x^2 /(i−(√2))) e^((i−(√2))x)  −(2/(i−(√2))) {  (x/(i−(√2))) e^((i−(√2))x)  −(1/(i−(√2))) e^((i−(√2))x) }  =(x^2 /(i−(√2))) e^((i−(√2))x)  −((2x)/((i−(√2))^2 )) e^((i−(√2))x)  +(2/((i−(√2))^2 )) e^((i−(√2))x)  =...  v_2 =∫ (W_2 /W)dx =−(1/(2(√2)))∫ x^2  sinx e^(x(√2))  =....(we follow the same way)  ⇒y_p =u_1 v_1  +u_2 v_2  and general solution is y =y_h  +y_p

(he)y2y=0r22=0r=+2yh=ae2x+be2x=au1+bu2W(u1,u2)=|ex2ex22ex22e2x|=22W1=|oex2x2sinx2e2x|=x2sinxex2W2=|ex202ex2x2sinx|=x2sinxex2v1=W1Wdx=122x2sinxex2dx=122Im(x2eixx2dx)x2e(i2)xdx=byparts1i2x2e(i2)x1i22xe(i2)x[dx=x2i2e(i2)x2i2{xi2e(i2)x1i2e(i2)x}=x2i2e(i2)x2x(i2)2e(i2)x+2(i2)2e(i2)x=...v2=W2Wdx=122x2sinxex2=....(wefollowthesameway)yp=u1v1+u2v2andgeneralsolutionisy=yh+yp

Answered by mathmax by abdo last updated on 05/Jun/20

let use Laplace transform  (e) ⇒L(y^((2)) )−2L(y) =L(x^2  sinx) ⇒  x^2 L(y)−y(0)−y^′ (0)−2L(y)=L(x^2  sinx) ⇒  (x^2 −2)L(y)=1+L(x^2  sinx) we have L(x^2 sinx) =∫_0 ^∞  t^2  sint e^(−xt)  dt  =Im(∫_0 ^∞  t^2 e^(it−xt)  dt) but ∫_0 ^∞  t^2  e^((i−x)t) dt =_(by parts)   [(t^2 /(i−x))e^((i−x)t) ]_0 ^∞ −∫_0 ^∞  ((2t)/(i−x)) e^((i−x)t) dt =−(2/(i−x)) ∫_0 ^∞  t e^((i−x)t)  dt  =(2/(x−i)){  [(t/(i−x)) e^((i−x)t) ]_0 ^∞  −∫_0 ^∞  (1/(i−x))e^((i−x)t)  dt}  =(2/((x−i)^2 )) ×[(1/(i−x)) e^((i−x)t) ]_0 ^∞  = (2/((x−i)^3 )) =((2(x+i)^3 )/((x^2  +1)^3 )) =((2(x^3  +3x^2 i −3x−i))/((x^2  +1)^3 ))  =((2x^3  +6x^2 i−6x −2i)/((x^2  +1)^3 )) ⇒L(x^2 sinx) =((6x^2 −2)/((x^2  +1)^3 ))  (e)⇒(x^2 −2)L(y) =1+((6x^2 −2)/((x^2  +1)^3 )) ⇒L(y) =(1/(x^2 −2)) +((6x^2 −2)/((x^2 −2)(x^2 +1)^3 )) ⇒  y(x) =L^(−1) ((1/(x^2 −2))) +L^(−1) (((6x^2 −2)/((x^2 −2)(x^2 +1)^3 )))  (1/(x^2 −2)) =(1/(2(√2)))((1/(x−(√2)))−(1/(x+(√2)))) ⇒L^(−1) ((1/(x^2 −2))) =(1/(2(√2)))(e^(x(√2))  −e^(−x(√2)) )  let decompose F(x) =((6x^2 −2)/((x^2 −2)(x^2  +1)^3 ))  F(x) =(a/(x−(√2))) +(b/(x+(√2))) +((a_1 x +b_1 )/(x^2  +1)) +((a_2 x +b_2 )/((x^2  +1)^2 )) +((a_3 x +b_3 )/((x^2  +1)^3 ))  ...be continued...

letuseLaplacetransform(e)L(y(2))2L(y)=L(x2sinx)x2L(y)y(0)y(0)2L(y)=L(x2sinx)(x22)L(y)=1+L(x2sinx)wehaveL(x2sinx)=0t2sintextdt=Im(0t2eitxtdt)but0t2e(ix)tdt=byparts[t2ixe(ix)t]002tixe(ix)tdt=2ix0te(ix)tdt=2xi{[tixe(ix)t]001ixe(ix)tdt}=2(xi)2×[1ixe(ix)t]0=2(xi)3=2(x+i)3(x2+1)3=2(x3+3x2i3xi)(x2+1)3=2x3+6x2i6x2i(x2+1)3L(x2sinx)=6x22(x2+1)3(e)(x22)L(y)=1+6x22(x2+1)3L(y)=1x22+6x22(x22)(x2+1)3y(x)=L1(1x22)+L1(6x22(x22)(x2+1)3)1x22=122(1x21x+2)L1(1x22)=122(ex2ex2)letdecomposeF(x)=6x22(x22)(x2+1)3F(x)=ax2+bx+2+a1x+b1x2+1+a2x+b2(x2+1)2+a3x+b3(x2+1)3...becontinued...

Commented by mathmax by abdo last updated on 05/Jun/20

error at line 2 →x^2  L(y)−xy(o)−y^′ (0)−2L(y)=L(x^2 sinx)  but this dont change the result because y(0)=0!

erroratline2x2L(y)xy(o)y(0)2L(y)=L(x2sinx)butthisdontchangetheresultbecausey(0)=0!

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