dy/dx=(1+Lnx)y and if y=1 when x=1, then y=?
i know i have to find the integral by multiplying dx to the other side and bringing y to left.
and i got Ln(y)+ C = x + 1/x + C
what do i do next.
ln(y)=x+1/x+c
e^(ln(y))=e^(x+1/x+c), so
y= e^(x+1/x+c)
y= Ce^(x+1/x)
Great job so far! After reaching the equation ln(y) = x + 1/x + C, you correctly applied the exponential function to both sides by taking the base e^( ) of both sides.
Remember that e^(ln(y)) simplifies to just y, so you can rewrite the equation as y = e^(x + 1/x + C).
Since C is an arbitrary constant, you can combine it with e^(x + 1/x) to form a new constant, let's call it C'.
Therefore, the final equation for y is y = C'e^(x + 1/x).
Now based on the initial condition y = 1 when x = 1, you can substitute these values into the equation and solve for the value of C'.
So, when x = 1, y = 1:
1 = C'e^(1 + 1/1)
1 = C'e^2
Now, solve for C':
C' = 1/e^2
And the final equation for y is y = (1/e^2)e^(x + 1/x).