For each of the exercises given below, verify that the given function (implicit or explicit) is a solution of the corresponding differential equation.(i) `y=a e^x+b e^(-x)+x^2` : `x(d^2y)/(dx^2)+2y(dy)/(dx)-x y+x^2-2=0`(ii) `y=e^x(acosx+bsin

Explanation

`(i)` Given,
`x(d^(2)y)/(dx^(2))+2(dy)/(dx)-xy+x^(2)-2=0`………`(1)`
and `y=ae^(x)+be^(-x)+2x`
`implies (dy)/(dx)=ae^(x)-be^(-x)+2x`
`implies (d^(2)y)/(dx^(2))-ae^(x)+be^(-x)+2`
Now, put the values of `y`, `(dy)/(dx)` and `(d^(2)y)/(dx^(2))` in equation `(1)`
`L.H.S =x(d^(2)y)/(dx^(2))+2(dy)/(dx)-xy+x^(2)-2`
`=x(ae^(x)+be^(-x)+2)+2(ae^(x)-be^(-x)+2x)-x(ae^(x)+be^(-x)+x^(2))+x^(2)-2`
`=(axe^(x)+bxe^(-x)+2x)+(2ae^(x)-2be^(-x)+4x)-(axe^(x)+bxe^(-x)+x^(3))+x^(2)-2`
`= axe^(x)+bxe^(-x)+2x+2ae^(x)-2be^(-x)+4x-axe^(x)-bxe^(-x)-x^(3)+x^(2)-2`
`=2ae^(x)-2be^(-x)-x^(3)+x^(2)+6x-2ne0`
`R.H.S ne L.H.S`
Therefore, given function is not the solution of the given differential equation.
`(ii)` Given `y=e^(x)(a cosx+bsinx)`
`impliese^(-x)y=acosx+bsinx`.........`(1)`
`implies e^(-x)(dy)/(dx)-ye^(-x)=-asinx+bcosx`
`=e^(-x)(d^(2)y)/(dx^(2))-(dy)/(dx)e^(-x)-(-ye^(-x)+e^(-x)(dy)/(dx))`
`=-a cosx-bsinx`
`implies e^(-x)(d^(2)y)/(dx^(2))-2e^(-x)(dy)/(dx)+ye^(-x)-e^(-x)(dy)/(dx)`
`=-(a cos x-bsinx)`
`implies e^(-x)(d^(2)y)/(dx^(2))-2e^(-x)(dy)/(dx)+ye^(-x)=-ye^(-x)` [from equation `(1)` ]
`implies e^(-x)(d^(2)y)/(dx^(2))-2e^(-x)(dy)/(dx)+2ye^(-x)=0`
`implies e^(-x)((d^(2)y)/(dx)-2(dy)/(dx)+2y)=0`
`implies (d^(2)y)/(dx^(2)-2(dy)/(dx)+2y=0`
Therefore, given function is the solution of the given differential equation.
`(iii)` Given, `y=x sin3x`
`implies (dy)/(dX)=x(d)/(dx)(sin3x)+sin3x(d)/(dx)(x)`
`(dy)/(dx)=x cos3xxx3+sin3x`
`=3xcox3x+sin3x`
`implies (d^(2)y)/(dx^(2))=3[x(d)/(dx)cos3x+cos3x(d)/(dx)(x)]+(d)/(dx)(sin3x)`
`implies (d^(2)y)/(dx^(2))=3[x(-sin3xx x3)+cos3x]+cos3xx x3`
`implies (d^(2)y)/(dx^(2))=-9xsin3x+3cos3x+3cos3x`
`implies (d^(2)y)/(dx^(2))=-9xsin3x+6cos3x`
`implies (d^(2)y)/(dx^(2))=-9y+6cos3x`
[From equation `(1)`, `y=xsin3x`]
`implies (d^(2)y)/(dx^(2))+9y-6cos3x=0`
Therefore, the given function is the solution of the given differential equation.
`(iv)` Given, `x^(2)=2y^(2)logy`
`implies2x=2(y^(2)xx(1)/(y)(dy)/(dx)+logyxx2y(dy)/(dx))`
`implies 2x=2(y+2ylogy)(dy)/(dx)`
`impliesx=(y+2ylogy)(dy)/(dx)`
`impliesxy=(y^(2)+2y^(2)logy)(dy)/(dx)`
`implies xy=(y^(2)+x^(2))(dy)/(dx)`
`implies (x^(2)+y^(2))(dy)/(dx)-xy=0`
Therefore, the given function is the solution of given differential equation.