Question

Let $y=y(x)$ be the solution of the differential equation $x^3 d y+(x y-1) d x=0, x>0$, $y\left(\frac{1}{2}\right)=3- e$ Then $y(1)$ is equal to

• A. $2-e$
• B. e
• C. 1
• D. 3

Answer 1

Answer: (C)

1

Answer 2

dxdy​=x31−xy​=x31​−x2y​
dxdy​+x2y​=x31​
If =e∫x21​dx=e−x1​
y⋅e−x1​=∫e−x1​⋅x31​dx( put −x1​=t)
y⋅e−x1​=−∫et⋅tdt
y=x1​+1+Cex1​
Where C is constant
Put x=21​
3−e=2+1+Ce2
C=−e1​
y(1)=1