Question

solve (d²-1)y= coshx cosx
​

Answer 1

Answer:

(D2+1)y=x.cos(x) , so auxiliary equation is ⟹D2+1=0

D=±i⟹ yC.F=C1cos(x)+C2sin(x)

Now to the P.I , yP.I=x.cos(x)D2+1=12.e−ixD2+1.x+12.eixD2+1.x

Now use this..

yPI=eax.V(x)f(D)=eax.V(x)f(D+a)

so we’ll have, yP.I=12.e−ix(D−i)2+1.x+12.eix(D+i)2+1.x

⟹12.e−ixxD(D−2i)+12.eixxD(D+2i)

⟹12.e−ix(1D.(D−2i)−1.x)+12.eix(1D.(D+2i)−1.x)

now expanding (D−2i)−1 and (D+2i)−1

⟹12.e−ix[1D(i2+14D)x]+12.eix[1D(−i2+14D)x]

where D→ddx and 1D→∫

⟹12.e−ix[i4.x2+x4]+12.eix[−i4x2+x4]

⟹x24(eix−e−ix2i)+x4(eix+e−ix2)

⟹yP.I=x24sin(x)+x4cos(x)

and the total solution is

⟹y(x)=yC.F+yP.I=C1cos(x)+C2sin(x)+x24sin(x)+x4cos(x)

Step-by-step explanation:

please mark as brain list

Answer 2

Answer:

Heya Army

Here is Your Answer Below

Step-by-step explanation:

(D2+1)y=x.cos(x) , so auxiliary equation is ⟹D2+1=0

D=±i⟹ yC.F=C1cos(x)+C2sin(x)

Now to the P.I , yP.I=x.cos(x)D2+1=12.e−ixD2+1.x+12.eixD2+1.x

Now use this..

yPI=eax.V(x)f(D)=eax.V(x)f(D+a)

so we’ll have, yP.I=12.e−ix(D−i)2+1.x+12.eix(D+i)2+1.x

⟹12.e−ixxD(D−2i)+12.eixxD(D+2i)

⟹12.e−ix(1D.(D−2i)−1.x)+12.eix(1D.(D+2i)−1.x)

now expanding (D−2i)−1 and (D+2i)−1

⟹12.e−ix[1D(i2+14D)x]+12.eix[1D(−i2+14D)x]

where D→ddx and 1D→∫

⟹12.e−ix[i4.x2+x4]+12.eix[−i4x2+x4]

⟹x24(eix−e−ix2i)+x4(eix+e−ix2)

⟹yP.I=x24sin(x)+x4cos(x)

and the total solution is

⟹y(x)=yC.F+yP.I=C1cos(x)+C2sin(x)+x24sin(x)+x4cos(x)

Hope it will be Helpful

Keep Smiling

Be Happy Forever

and I Purple You Army