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))`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))`

The solution of the differential equation `xdy/dx+2y=x^(2),(xne0)` wit

1.	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))`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» Correct Answer - (a) 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)`

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