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Solve the following initial-value problems by using integrating factors.$$x y^{\prime}=y-3 x^{3}, y(1)=0$$
$y=-\frac{3 x^{3}}{2}+\frac{3}{2} x$
Calculus 2 / BC
Chapter 4
Introduction to Differential Equations
Section 5
First-order Linear Equations
Differential Equations
Oregon State University
Baylor University
University of Michigan - Ann Arbor
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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.
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