Question

Let y(x) be the solution of the initial value problemx^2y"+xy'+y=x and y(1)=y'(1)=1.then the value of y(e^90°)
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Answer 1

Answer: The value of y(e^90°) is 2.90

Step-by-step explanation:

Step1: Finding homogeneous solution

x²y" + xy' + y = 0

Substitute $y = x^{m}$

$x^{m}$ [m(m-1) + m + 1] = 0

[m(m-1) + m + 1] = 0

m² + 1 = 0

m = i , -i

$Y_h$ = C.sin(lnx) + D.cos(lnx)

Step 2: Finding particular solution

x²y" + xy' + y = x

Substitute y = Ax+B

0 + Ax + Ax+B = x

2Ax + B = x

Therefore,

A = 0.5 , B = 0

$Y_p$ = x/2

Step 3: Writing general solution

y = $Y_h$ + $Y_p$

y = C.sin(lnx) + D.cos(lnx) + x/2

Step 4: Substituting initial values to find C and D

y(1) = 1

Therefore,

1 = 0 + D + 0.5

D = 0.5

y’(x) = C.(cos(lnx))/x - D.(sin(lnx))/x + 1/2

y’(1) = 1

Therefore,

1 =  C + 0 + 0.5

C = 0.5

y = ( sin(lnx) + cos(lnx) + x ) / 2

At  x = e^90° =  e^(/2) =  4.81,

y(x) =  ( sin(ln e90°) + cos(ln  e90° ) +4.81 ) /2

y(x) = ( 1 + 0 + 4.81) /2

y(x) = 2.905

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Answer 2

Answer:

Step-by-step explanation:

$x^{2} y''+ xy'=0\\substitute y=x^{m}\\ x^{m}[m(m-1)+m+1]=0\\ [m(m-1)+m+1=0]\\m^{2}+1=0\\ m=1, -i\\yh=C.sin(inx)+D,cos(inx)$

step number 2

find particular solution.

$x^{2} y''+xy'=x\\substitute y=Ax+B\\O+Ax+Ax+B\\2Ax+B=x\\therefore,\\A=0.5, B=0\\yp =x/2$

step number 3

writing general solution.

$y=yh+yp\\y=C.sin(inx)+D.cos(inx)+x/2$

step number 4

substituting initial values to find C and D

$y(1)=1\\therefore,\\1=0+D+0.5\\\\y'(x)=C.(cos(inx))/x- D.(sin(inx))/x+1/2\\therefore\\1=C+0.5+0\\C=0.5$

y(x)=2,905