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I'm having trouble starting this problem. I need help on determining the integrating factor by first setting up in standard form...

$y' + p(x)y = f(x) $

Solve for IVP.

$$yy' + 3t^2 - 2 = 0$$ $$y(-1) = 2$$

Seeing the "$yy'$" together is throwing me off.

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$$yy'+3t^{ 2 }-2=0\\ { \left( { y }^{ 2 } \right) }^{ \prime }=4-6{ t }^{ 2 }\\ { y }^{ 2 }=4t-2{ t }^{ 3 }+C$$ as we are given $\\ y(-1)=2$ so $$4=-4+2+C\quad \\ C=6$$

$${ y }^{ 2 }=4t-2{ t }^{ 3 }+6 $$

  • 0
    From the first step to the second step, How did you know to multiply by 2? Is it because it's what is given in $y(-1) = 2$? – 2017-02-12
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    no, because of ${ \left( { y }^{ 2 } \right) }^{ \prime }=2yy'$ – 2017-02-12