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 2 years ago
Find the general solution of the given differential equations. Give the largest interval I over which the general solution is defined. Determine whether there are any transient terms in the general solution.
y' + 2xy = (x^3)
ydx = (y(e^y)  2x)dy
(dP)/(dt) + 2tP = P + 4t  2
 2 years ago
Find the general solution of the given differential equations. Give the largest interval I over which the general solution is defined. Determine whether there are any transient terms in the general solution. y' + 2xy = (x^3) ydx = (y(e^y)  2x)dy (dP)/(dt) + 2tP = P + 4t  2

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hellow
 2 years ago
Best ResponseYou've already chosen the best response.0The last one is separable. dP/dt = P2tP + 4t2 dP/dt = P(12t) + 2(2t1) = P(12t) 2(12t) dP/dt = (P2)(12t) 1/(P2)dP = (12t)dt integrate both sides lnP2=tt^2+c This solution is defined for all P except P= 2 (since ln(0) is not defined). But we can check the solution P= 2. Plugging that into the original ODE we get 0 +2t(2)= 2+4t2 4t=4t. So P=2 also works. So, although the given solution is not defined for P=2, there is a solution when P=2. You could try integrating factors for the first problem, but there might be an easier way. I am not sure what to try for the middle problem.

hellow
 2 years ago
Best ResponseYou've already chosen the best response.0I also realized that the middle one is linear in terms of x. You have \[y \frac{ dx }{ dy}=y e^{y}2x\] or \[y \frac{ dx }{ dy}+2x=y e^{y}\] or\[\frac{ dx }{ dy}+2x/y=e^{y}\]which is linear with respect to x and its derivatives. Thus it can be solved with integrating factors. Once you calculate your integrating factor, I think you will find it is y^2. Thus\[(xy ^{2})'=e ^{y}y ^{2}\]\[xy ^{2}=\int\limits_{}^{}e ^{y}y ^{2}dy\]\[xy ^{2}=y ^{2}e ^{y}2y e ^{y}+2e ^{y}+c\]

hellow
 2 years ago
Best ResponseYou've already chosen the best response.0The first problem can be solved in a similar way. It is linear with respect to y and its derivatives. The integration factor is e^(x^2), which means that the right hand side becomes \[\int\limits_{?}^{?}x ^{3}e ^{x ^{2}}dx\] If you let u=x^2, then the integral becomes much easier. Using the table method for inegration by parts I came up with \[x ^{2}y=1/2(x ^{2}e ^{x ^{2}}e ^{x ^{2}})+c\]
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