• anonymous
find the general solution to the first order differential equation: 2ty'+4y=3 i know this is really basic, but i can not figure out the correct steps.
  • Stacey Warren - Expert
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  • katieb
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  • anonymous
The general method is called the "Method of Integrating Factors." The idea is to multiply both sides by some "integrating factor" so that the Left Hand Side becomes the derivative of a product of y with some other function. This is because if you differentiate f(x)*y(x) with respect to x, then you get f*y’ + f’y. So we’d get some function times y’ added to some function times y, which is a similar form to the Left Hand Side of these linear ODE’s. Normally when using this method, especially if you can’t spot what to multiply both sides by, you first want the coefficient on y’ to be 1. So dividing by 2t, we get y’ + (2/t) y = 3/(2t). So we want to multiply both the Left and Ride sides by some function, say f, so that the Left Hand Side becomes the derivative of a product of y with some other function. What should that f be? Suppose we had that f, and multiplied both sides by f. Then the Left Hand Side becomes fy’ + (2f/t) y. Now remember that we want this to match up with f*y’ + f’y. So we want f’ to be equal to 2f/t. Setting f’ = 2f/t, we want to solve for f(t). That’s easy, f’/f = 2/t. So ln f = 2 ln t. So f = t^2. (We don’t need to worry about constants here since they’ll be absorbed in the final answer.) Therefore, if we multiply both sides of y’ + (2/t) y = 3/(2t) by t^2, we’d find that the Left Hand Side would become the derivative of (t^2)*y(t). The Right Hand Side would become 3t/2. Since the Left and Right Sides are equal, this means that 3t/2 is the derivative of (t^2)*y(t) with respect to t. Integrating both sides with respect to t, we have that (3t^2)/4 + C = (t^2)*y(t). So y(t) = 3/4 + C/t^2.

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