SOLUTION: 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 d

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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

Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
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
y%5E2%2Aexpr%28dx%2Fdy%29%22%22=%22%22x%2B1

Multiply both sides by dy

y%5E2%2Adx%22%22=%22%22%28x%2B1%29dy

Divide every term by y²(x+1)

dx%2F%28x%2B1%29%22%22=%22%22dy%2Fy%5E2

dx%2F%28x%2B1%29%22%22=%22%22y%5E%28-2%29dy

Now we integrate every term

int%28dx%2F%28x%2B1%29%29%22%22=%22%22int%28y%5E%28-2%29dy%29

ln%28x%2B1%29%22%22=%22%22y%5E%28-1%29%2F%28-1%29%22%22%2B%22%22C

ln%28x%2B1%29%22%22=%22%22-1%2Fy%22%22%2B%22%22C

y%2Aln%28x%2B1%29%22%22=%22%22-1%22%22%2B%22%22C%2Ay

We solve for y:

y%2Aln%28x%2B1%29-C%2Ay%22%22=%22%22-1

y%28ln%28x%2B1%29-C%29%22%22=%22%22-1

y%22%22=%22%22-1%2F%28ln%28x%2B1%29-C%29

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%22%22=%22%22-1%2F%28%28ln%281%2B1%29-C%29%29

We solve for C

-1%22%22=%22%22-1%2F%28ln%282%29-C%29

1%22%22=%22%221%2F%28ln%282%29-C%29

ln%282%29-C%22%22=%22%221

ln%282%29-1%22%22=%22%22C

We substitute that value for C in the general solution:

y%22%22=%22%22-1%2F%28ln%28x%2B1%29-%28ln%282%29-1%29%29

y%22%22=%22%22-1%2F%28ln%28x%2B1%29-ln%282%29%2B1%29

Edwin