- #1
prace
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I am asked to solve this DE with the initial condition of y(1) = 1.
[tex](x+y)^2dx + (2xy + x^2-1)dy = 0[/tex]
So, after working the problem out, I came to this as an answer:
[tex]F(x,y)=\frac{1}{3}x^3 + x^2y + xy^2-y[/tex]
My question is what do I do with the initial condition. I assume that I am just suppossed to plug 1 in somewhere, but the syntax of the initial condition does not seem very intuitive to me. What does y(1) = 1 mean?
Thank you
[tex](x+y)^2dx + (2xy + x^2-1)dy = 0[/tex]
So, after working the problem out, I came to this as an answer:
[tex]F(x,y)=\frac{1}{3}x^3 + x^2y + xy^2-y[/tex]
My question is what do I do with the initial condition. I assume that I am just suppossed to plug 1 in somewhere, but the syntax of the initial condition does not seem very intuitive to me. What does y(1) = 1 mean?
Thank you