James10492
Junior Member
- Joined
- May 17, 2020
- Messages
- 50
Hi there, I have just come across a couple of differentials I cannot solve. I have applied the methods I usually use for solving differential equations but frustratingly my solutions do not match the ones in my book. Please could someone point out what I am not doing right?
Problem one:
given that y = 6 and x = .5, find the particular solution:
[math](1-x^2)\frac{dy}{dx} = xy + y[/math]
[math](1-x^2)\frac{dy}{dx} = y(x+1)[/math]
[math]\frac{dy}{y} = \frac{(1+x)}{(1+x)(1-x)} dx[/math]
[math]\int\frac{dy}{y} = \int\frac{1}{(1-x)}[/math]
[math]\ln{y} = \ln{(1-x)} + \ln{k}[/math]
[math]y = k(1-x)[/math]
[math]6 = k(1-.5) \\ 6= .5k \\ k = 12[/math]
so the particular solution is 12(1-x), right?
Problem two:
find the particular solution for y=2, x =0:
[math](1+x^2)\frac{dy}{dx} = x - xy^2 \\ (1+x^2)\frac{dy}{dx} = x(1 - y^2) \\ \frac{dy}{1-y^2} = \frac{x}{1+x^2}dx \\[/math]
writing the left-hand side as partial fractions:
[math]\int\frac{1}{2(1+y)} + \frac{1}{2(1-y)} dy = \int\frac{x}{1+x^2} dx \\ \frac{1}{2} \ln{(1+y)} - \frac{1}{2}\ln(1-y) = \frac{1}{2}\ln(1+x^2) + C \\ \ln{(1+y)} - \ln(1-y) = \ln(1+x^2) + \ln {k}\\ \frac{(1+y)}{(1-y)} = k(1+x^2) \\ -\frac{3}{2} = k[/math]
I think something is probably wrong because now it is difficult to write y = f(x)
Problem one:
given that y = 6 and x = .5, find the particular solution:
[math](1-x^2)\frac{dy}{dx} = xy + y[/math]
[math](1-x^2)\frac{dy}{dx} = y(x+1)[/math]
[math]\frac{dy}{y} = \frac{(1+x)}{(1+x)(1-x)} dx[/math]
[math]\int\frac{dy}{y} = \int\frac{1}{(1-x)}[/math]
[math]\ln{y} = \ln{(1-x)} + \ln{k}[/math]
[math]y = k(1-x)[/math]
[math]6 = k(1-.5) \\ 6= .5k \\ k = 12[/math]
so the particular solution is 12(1-x), right?
Problem two:
find the particular solution for y=2, x =0:
[math](1+x^2)\frac{dy}{dx} = x - xy^2 \\ (1+x^2)\frac{dy}{dx} = x(1 - y^2) \\ \frac{dy}{1-y^2} = \frac{x}{1+x^2}dx \\[/math]
writing the left-hand side as partial fractions:
[math]\int\frac{1}{2(1+y)} + \frac{1}{2(1-y)} dy = \int\frac{x}{1+x^2} dx \\ \frac{1}{2} \ln{(1+y)} - \frac{1}{2}\ln(1-y) = \frac{1}{2}\ln(1+x^2) + C \\ \ln{(1+y)} - \ln(1-y) = \ln(1+x^2) + \ln {k}\\ \frac{(1+y)}{(1-y)} = k(1+x^2) \\ -\frac{3}{2} = k[/math]
I think something is probably wrong because now it is difficult to write y = f(x)