anonymous
  • anonymous
how to solve dy/dx=x+y,y(0)=1
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
apples
  • apples
Noting that this is a linear ordinary differential equation, we can rewrite it in the form \[\frac{dy}{dx} + p(x) \cdot y = g(x)\] where \[p(x) = -1, g(x) = x\]Finding the integration factor u(x) is trivial, as \[u(x) = e^{\int p(x) dx} = e^{-x}\] Multiplying both sides by the integration factor, we arrive at \[e^{-x} \frac{dy}{dx} + e^{-x} \cdot -y = e^{-x} \cdot x\] The left side may be rewritten as \[(e^{-x} \cdot y)' = e^{-x} \cdot x\] and integrating both sides, we find that \[\int (e^{-x} \cdot y)' dx = \int e^{-x} x dx\] which is \[e^{-x} \cdot y + c_1 = -e^{-x} (x+ 1) + c_2\] for some c_1, c_2. Letting \[c = c_2 - c_1\] we may rewrite the differential equation as \[e^{-x} y = -e^{-x} (x + 1) + c\] Solving for y, we arrive at \[y = \frac{-e^{-x} (x+ 1) + c}{e^{-x}} = c e^{x} - x - 1\] which is our general solution. Solving for c we find that \[y(0) = c \cdot e^{0} - 0 - 1 = c - 1 = 1\] thus c equals 2.

Looking for something else?

Not the answer you are looking for? Search for more explanations.