Question 851663
Calc issues

Determine the General and Particular solutions of this First Order Differential equation by direct integration or by separating the variables as appropriate

y^2 dx/dy= x + 1	given that when  y=-1, x=1
<pre>
{{{y^2*expr(dx/dy)}}}{{{""=""}}}{{{x+1}}}

Multiply both sides by dy

{{{y^2*dx}}}{{{""=""}}}{{{(x+1)dy}}}

Divide every term by y²(x+1)

{{{dx/(x+1)}}}{{{""=""}}}{{{dy/y^2}}}

{{{dx/(x+1)}}}{{{""=""}}}{{{y^(-2)dy}}}

Now we integrate every term

{{{int(dx/(x+1))}}}{{{""=""}}}{{{int(y^(-2)dy)}}}

{{{ln(x+1)}}}{{{""=""}}}{{{y^(-1)/(-1)}}}{{{""+""}}}{{{C}}}

{{{ln(x+1)}}}{{{""=""}}}{{{-1/y}}}{{{""+""}}}{{{C}}}

{{{y*ln(x+1)}}}{{{""=""}}}{{{-1}}}{{{""+""}}}{{{C*y}}}

We solve for y:

{{{y*ln(x+1)-C*y}}}{{{""=""}}}{{{-1}}}

{{{y(ln(x+1)-C)}}}{{{""=""}}}{{{-1}}}

{{{y}}}{{{""=""}}}{{{-1/(ln(x+1)-C)}}}

That's the general solution.

Now we substitute: y=-1, x=1 to find which particular 
solution is the equation for a graph of a curve that
goes through the point (1,-1)

{{{-1}}}{{{""=""}}}{{{-1/((ln(1+1)-C))}}}

We solve for C

{{{-1}}}{{{""=""}}}{{{-1/(ln(2)-C)}}}

{{{1}}}{{{""=""}}}{{{1/(ln(2)-C)}}}

{{{ln(2)-C}}}{{{""=""}}}{{{1}}}

{{{ln(2)-1}}}{{{""=""}}}{{{C}}}

We substitute that value for C in the general solution:

{{{y}}}{{{""=""}}}{{{-1/(ln(x+1)-(ln(2)-1))}}}

{{{y}}}{{{""=""}}}{{{-1/(ln(x+1)-ln(2)+1)}}}

Edwin</pre>