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Ch2.1- Linear First Order Integrating F Previous Problem Problem List Next Problempoint)Solve the initial value problem y Xty =x-2 y(x)Y(1) = 3...

Question

Ch2.1- Linear First Order Integrating F Previous Problem Problem List Next Problempoint)Solve the initial value problem y Xty =x-2 y(x)Y(1) = 3

Ch2.1- Linear First Order Integrating F Previous Problem Problem List Next Problem point) Solve the initial value problem y Xty =x-2 y(x) Y(1) = 3



Answers

Solve the first-order linear differential equation.
$y^{\prime}-3 x^{2} y=e^{x^{3}}$

Section four out five number 2 43 were dealing with first order linear, ordinary differential equations. If I could get this in standard form, that is why prime plus some function of X times y is equal to some other function of X. I know that I can find the integrating factor and saw this different equation. So to get this in standard form, that's going to be why prime minus one over X, why it's equal to minus three x cubed over X, which is just minus three x squared. And so the integrating factor is going to be e raised to the anti derivative of minus one over X dx. So this is e to the minus natural log of X E to the natural log off extra the minus one, which is just one over tax. So that is my integrating factor. So I can use that to rewrite this equation by by multiplying everything by one of her acts. So I get why over X prime is equal to minus three x squared over X, which is just minus three X, and now the key to solving this will be to integrate both sides on the left side. I simply get why over X and on the right, I get minus three x squared over to plus a constant of integration. And then to solve this for why you multiply everything by x so minus three X cubed over to plus C X and now you have your boundary condition was Why of one is zero. So why of 10? That means that zero is equal to minus three halves plus C, which tells me that C is equal to three halves. So the final solution here is why is equal to minus three have X cube plus three halves X, and that is our solution.

So a section 4.5 problem to 41. We're dealing with a linear, ordinary first order differential equation, and we have a boundary condition. 50 is equal to three on. So what we need to do is just to put this in standard form. And we say, Well, it's already in standard forms. A standard form is why prime some function of X times. Why is equal to some other function of X so I could find the integrating factor is going to be e and so the coefficient of y just one. So this is going to be e to the X is the integrating factor. So that tells me I can rewrite this differential equation as why into the X prime is equal to X e to the X, Then the solution would be to integrate both sides of this equation. So in integrating both sides of this equation, I end up with why e to the X is equal to either the ex text minus one plus a constant of integration then to solve for why it's just a divide by either the x o why is equal to X minus one plus c e to the minus X. Then you have the boundary condition that why of zero 2nd 3 that tells me that to solve for C, I'm gonna have a three. It's a people to zero minus one. See? And then each of the zeros just simply one. So that tells me that C is equal to four. So the final answer here is why is equal to X minus? One was for E to the minus X, and that is the exact solution to this differential question.

Okay, whatever. That X Y time is your ex to party. It's you minus three. Why? Over X? Okay, so we can rewrite this equation as y prime. So this is a crime here is equal to Hmm. Let's see. We need the while for here. So we have one of our exit three. Why? Over X squared. Negative plus one over X Cute. Well, it's a ad. Dis portion two left. So they get wide prime is equal to Or why? If I'm close three. Why over X squared is one over X cube. Who? We have it in the form we want. This was, you know, here is a of X. This is be of X. So based on this or integrating background of ex, you could be in a girl. Three over X squared X. Yeah, that gives me each of them. They go three. Excellent. One. Okay, Now multiply E living three excellent ones to this and get that the You got a three over X y prime three over X squared each in the three off. X Y is equal to one over X cubed each and your three ex. Okay, Now, if we interrogate both sides. What do we get? Well, we get that he made three over x times way physical to expose three over nine X, the agent and a good three over X pussy and are dividing by beaning three off. Except besides, get that. Why is he going to expose three over nine X? But see, you did it to be over X has a solution.

Hello, everyone. Today we are going to solve a problem. Number developed, you know, they given the education is wild eyes plus three by because you know the power three x so p off X equal to three and cue off its equals. You did about three years. So integrating factory workers into the power in the three d x with difficult o u to the power three x So you know the power three x into white equals integral into the power three x into you know the power three x the X, which is a cardio in the valued at about 66 de it's so they will get like eight of the power he was is equal to one by six areas to 66 plus sick so into the power big white Because if the power three x they were by six plus C in the par minus three X, that is a solution. Thank you


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