The solution of the differential equation `y^-8y^=0,` where `y(0)=1/8,y^(prime)(0)=0,y^(0)=1` , is (a) `( b ) (c) y=( d )1/( e )8( f ) (g)(( h ) (i) (j)(( k ) (l) e^(( m ) (n)8x (o))( p ))/( q )8( r ) (s)+x-( t )7/( u )9( v ) (w) (x))( y )` (z) (aa) `( b b ) (cc) y=( d d )1/( e e )8( f f ) (gg)(( h h ) (ii) (jj)(( k k ) (ll) e^(( m m ) (nn)8x (oo))( p p ))/( q q )8( r r ) (ss)+x+( t t )7/( u u )8( v v ) (ww) (xx))( y y )` (zz) (aaa) `( b b b ) (ccc) y=( d d d )1/( e e e )8( f f f ) (ggg)(( h h h ) (iii) (jjj)(( k k k ) (lll) e^(( m m m ) (nnn)8x (ooo))( p p p ))/( q q q )8( r r r ) (sss)-x+( t t t )7/( u u u )8( v v v ) (www) (xxx))( y y y )` (zzz) (d) None of these

Options

A. `y=1/8(e^(8x)/8+x-7/9)`
B. `y=1/8(e^(8x)/8+x+7/8)`
C. `y=1/8(e^(8x)/8-x+7/8)`
D. None of these

Answer

C

Explanation

`y^(''')/y^('')=8` or `logy^('')=8x+c`
When `x=0, y^('')=1` or `log1=0`. Thus, c=0.
`therefore y^(')=e^(8x)`
Integrating again, we get
`y^(')=e^(8x)/8+lambda`
When `x=0, y^(')(0)=0`
`therefore lambda=-1/8`
`therefore y^(')=e^(8x)/8-1/8`.
Integrate again. Then,
`y=e^(8x)/64-x/8+k`
Also, when `x=0,y=1/8`. Thus, `k=7/64`.
`therefore y=1/8(e^(8x)/8-x+7/8)`