The solution of the differential equation `xdy/dx+2y=x^(2),(xne0)` with y(1)=1, is (a) `y=x^(2)/4+3/(4x^(2))` (b) `y=x^(3)/5+1/(5x^(2))` (c) `y=3/4x^(2)+3/(4x^(2))` (d) `y=4/5x^(3)+1/(5x^(2))`

Options

A. (a) `y=x^(2)/4+3/(4x^(2))`
B. (b) `y=x^(3)/5+1/(5x^(2))`
C. (c) `y=3/4x^(2)+3/(4x^(2))`
D. (d) `y=4/5x^(3)+1/(5x^(2))`

Answer

A

Explanation

Given differential equation is
`xdy/dx+2y=x^(2),(xne0)`
`rArr dy/dx+(2/x)y=x,`
shich is a linear differential equation of the form
`dy/dx+Py=Q`
Here, `P=2/x` and Q=x
`therefore IF=e^(int2/x dx) =e^(2log x) =x^(2) `
Since, solution of the given differential equation is
`yxxIF=int(QxxIF)dx+C`
` therefore y(x^(2)) = int (x xxx^(2)) dx + C rArr yx^(2) =x^(4)/4=C`
` therefore y(1) =1,`so `1=1/4+C rArr C=3/4`
` therefore yx^(2) =x^(4) /4+3/4 rArr y = x^(2) /4+3/(4x^(2)`