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Question Number 97797 by abdomathmax last updated on 09/Jun/20

solve y^(′′) −y = x

solvey″−y=x

Answered by niroj last updated on 09/Jun/20

y^(′′) −y = x (D^2 −1)y= x A.E.,m^2 −1=0 m=+^− 1 CF= C_1 e^x +C_2 e^(−x) y_1 =e^x , y_2 =e^(−x) , Q=x W= determinant (((e^x e^(−x) )),((e^x −e^(−x) ))) W= −e^(−x) .e^x −e^(−x) .e^x =−2 PI= −y_1 ∫((y_2 Q)/W)dx +y_2 ∫((y_1 Q)/W)dx = −e^x ∫((e^(−x) x)/(−2))dx+ e^(−x) ∫((e^x x)/(−2))dx = (e^x /2){x.(−e^(−x) )−∫1.(−e^x )dx}−(e^(−x) /2){x.e^x −∫1.e^x dx} = (e^x /2)(−xe^(−x) +e^x )−(e^(−x) /2)(xe^x −e^x ) = (1/2)(−xe^(x−x) +e^(2x) −xe^(x−x) +e^(x−x) ) = (1/2)(2e^(2x) −2x)= e^(2x) −x ∴ y= C_1 e^x +C_2 e^(−x) +e^(2x) −x //.

y″−y=x(D2−1)y=xA.E.,m2−1=0m=+−1CF=C1ex+C2e−xy1=ex,y2=e−x,Q=xW=|exe−xex−e−x|W=−e−x.ex−e−x.ex=−2PI=−y1∫y2QWdx+y2∫y1QWdx=−ex∫e−xx−2dx+e−x∫exx−2dx=ex2{x.(−e−x)−∫1.(−ex)dx}−e−x2{x.ex−∫1.exdx}=ex2(−xe−x+ex)−e−x2(xex−ex)=12(−xex−x+e2x−xex−x+ex−x)=12(2e2x−2x)=e2x−x∴y=C1ex+C2e−x+e2x−x//.

Answered by mathmax by abdo last updated on 09/Jun/20

(he)→y^(′′) −y =0 →r^2 −1=0 ⇒r =+^− 1 ⇒y_h =αe^x +β e^(−x) =αu_1 +βu_2 W(u_1 ,u_2 ) = determinant (((u_1 u_2 )),((u_1 ^′ u_2 ^′ )))= determinant (((e^x e^(−x) )),((e^x −e^(−x) )))=−2 W_1 = determinant (((0 e^(−x) )),((x −e^(−x) )))=−xe^(−x) and W_2 = determinant (((e^x 0)),((e^x x)))=xe^x v_1 =∫ (w_1 /w)dx =∫ ((−xe^(−x) )/(−2))dx =(1/2)∫ xe^(−x) dx =(1/2){− xe^(−x) +∫ e^(−x) dx} =−(1/2)(x+1)e^(−x) v_2 =∫ (w_2 /w)dx =∫ ((xe^x )/(−2))dx =−(1/2) ∫ xe^x dx =−(1/2){ xe^x −e^x } =(1/2)(1−x)e^x so the particular solution is y_p =u_1 v_1 +u_2 v_2 =−(1/2)(x+1)+(1/2)(1−x) =−x the general solution is y(x) =y_h +y_p =αe^x +β e^(−x) −x

(he)→y″−y=0→r2−1=0⇒r=+−1⇒yh=αex+βe−x=αu1+βu2W(u1,u2)=|u1u2u1′u2′|=|exe−xex−e−x|=−2W1=|0e−xx−e−x|=−xe−xandW2=|ex0exx|=xexv1=∫w1wdx=∫−xe−x−2dx=12∫xe−xdx=12{−xe−x+∫e−xdx}=−12(x+1)e−xv2=∫w2wdx=∫xex−2dx=−12∫xexdx=−12{xex−ex}=12(1−x)exsotheparticularsolutionisyp=u1v1+u2v2=−12(x+1)+12(1−x)=−xthegeneralsolutionisy(x)=yh+yp=αex+βe−x−x

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

let solve it by laplace e⇒L(y^(′′) )−L(y) =L(x) ⇒x^2 L(y)−xy(o)−y^′ (0)−L(y) =L(x) ⇒ (x^2 −1)L(y) =xy(0)+y^′ (0)+L(x) we have L(x)=∫_0 ^∞ t e^(−xt) dt =[−(t/x) e^(−xt) ]_(t=0) ^∞ +(1/x)∫_0 ^∞ e^(−xt) dt =−(1/x^2 )[e^(−xt) ]_(t=0) ^∞ =(1/x^2 ) (also we can use L(x^n ) =((n!)/x^(n+1) )) e ⇒(x^2 −1)L(y) =xy(o)+y^′ (0)+(1/x^2 ) ⇒ L(y) =(x/(x^2 −1))y(0) +((y^′ (0))/(x^2 −1)) +(1/(x^2 (x−1))) ⇒y(x) =y(0)L^(−1) ((x/(x^2 −1)))+y^′ (0)L^(−1) ((1/(x^2 −1))) +L^(−1) ((1/(x^2 (x−1)))) we havef(x) (x/(x^2 −1)) =(1/2)x((1/(x−1))−(1/(x+1))) =(1/2)((x/(x−1))−(x/(x+1))) =(1/2)(((x−1+1)/(x−1))−((x+1−1)/(x+1)))=(1/2)((1/(x−1))+(1/(x+1))) ⇒ L^(−1) (f)=(1/2)e^x +(1/2) e^(−x) g(x) =(1/(x^2 −1)) =(1/2)((1/(x−1))−(1/(x+1)))⇒L^(−1) (g) =(1/2)e^x −(1/2)e^(−x) h(x) =(1/(x^2 (x−1))) =(a/x) +(b/x^2 ) +(c/(x−1)) b =−1 , c =1 lim_(x→+∞) xh(x) =0 =a+c ⇒a=−1 ⇒ h(x) =−(1/x)−(1/x^2 ) +(1/(x−1)) ⇒L^(−1) (h) =−1−x +e^x ⇒ y(x) =y(0){((e^x +e^(−x) )/2)} +y^′ (0){((e^x −e^(−x) )/2)} −1 −x+e^x =(((y(o)+y^′ (0))/2)+1)e^x +(((y(0)−y^′ (0))/2))e^(−x) −x−1

letsolveitbylaplacee⇒L(y″)−L(y)=L(x)⇒x2L(y)−xy(o)−y′(0)−L(y)=L(x)⇒(x2−1)L(y)=xy(0)+y′(0)+L(x)wehaveL(x)=∫0∞te−xtdt=[−txe−xt]t=0∞+1x∫0∞e−xtdt=−1x2[e−xt]t=0∞=1x2(alsowecanuseL(xn)=n!xn+1)e⇒(x2−1)L(y)=xy(o)+y′(0)+1x2⇒L(y)=xx2−1y(0)+y′(0)x2−1+1x2(x−1)⇒y(x)=y(0)L−1(xx2−1)+y′(0)L−1(1x2−1)+L−1(1x2(x−1))wehavef(x)xx2−1=12x(1x−1−1x+1)=12(xx−1−xx+1)=12(x−1+1x−1−x+1−1x+1)=12(1x−1+1x+1)⇒L−1(f)=12ex+12e−xg(x)=1x2−1=12(1x−1−1x+1)⇒L−1(g)=12ex−12e−xh(x)=1x2(x−1)=ax+bx2+cx−1b=−1,c=1limx→+∞xh(x)=0=a+c⇒a=−1⇒h(x)=−1x−1x2+1x−1⇒L−1(h)=−1−x+ex⇒y(x)=y(0){ex+e−x2}+y′(0){ex−e−x2}−1−x+ex=(y(o)+y′(0)2+1)ex+(y(0)−y′(0)2)e−x−x−1

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