## Chena804 3 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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1. hellow

The last one is separable. dP/dt = P-2tP + 4t-2 dP/dt = P(1-2t) + 2(2t-1) = P(1-2t) -2(1-2t) dP/dt = (P-2)(1-2t) 1/(P-2)dP = (1-2t)dt integrate both sides ln|P-2|=t-t^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+4t-2 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.

2. hellow

I 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$

3. hellow

The 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 $x ^{2}y=1/2(x ^{2}e ^{x ^{2}}-e ^{x ^{2}})+c$