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For the following problems, find the solution to the initial value problem.$$y^{\prime}=3 y-x+6 x^{2}, y(0)=-1$$

$y(x)=-2 x^{2}-x-\frac{1}{3}-\frac{2}{3} c^{3 x}$

Calculus 2 / BC

Chapter 4

Introduction to Differential Equations

Section 5

First-order Linear Equations

Differential Equations

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01:11

In mathematics, integratio…

06:55

In grammar, determiners ar…

02:22

Find a solution to the ini…

03:34

For the following problems…

05:03

Solve the given initial-va…

04:00

Solve the initial value pr…

04:24

01:24

Solve the initial-value pr…

03:57

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Solve the following initia…

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11:02

so Ford up five problem to 77 were asked to solve this first order linear, ordinary, different show equation. So the key to solving this will be to get it in standard form. So why is equal to back up? Why prime plus a function of X times y is equal to another function of X and then my integrating factor will be e to the anti derivative of P of X. So standard form for this guy's gonna be y prime minus three y is equal to minus X plus six x squared. So the integrating factor is going to be e to the minus three the X So this is going to be e to the minus three X. So this equation is going to be why he to the three acts Prime is equal to minus X e to the three x plus six x squared e to the three X and sorry about that. It's minus. So it's minus three x and minus three X was the integrating factor there. So now the key will be, then, just to integrate both sides of this equation. So on the left side, you'll have why e to the minus three x and then when you perform this integration on the right side, minus 1/3 you to the minus three x and you're left with this polynomial terms six x square plus three x plus one, plus a constant integration. So then to solve for why, why is going to be minus 1/3 six X squared plus three x plus one plus c e to the minus three x Excuse me plus three x when you multiply that by negative 1/3 6 X square plus three x plus one ce to the three X that we have an initial condition y of zero is negative one. So if y zero equals negative one, then I get negative. One is equal to minus 1/3 zero plus zero plus one plus c and then e to the zeros just simply one. So this is minus one equal minus 1/3 plus see therefore C is equal to minus 2/3. So this final answer becomes why equal minus 1/3 six x squared plus three x plus one minus 2/3 e to them plus three x And that's the final answer

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