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## Homework Statement

Find the solution to:

y' = x.y.cos(x^2)

## Homework Equations

Integration by Parts method.

## The Attempt at a Solution

*Step 1*

(dy/dx).(1/y) = x.cos(x

^{2})

(1/y) dy = x.cos(x

^{2}) dx

*Step 2*

Integrate both sides.

ln|y| = integratal of [ x.cos(x

^{2}) dx ]

*Step 3*

Using integration by parts...

u = cos(x

^{2}) => du = -2x.sin(2^x)

dv = x => v = 1.dx

*Step 4*

Subbing back in...

ln|y| = u.v - integral of v.du = cos(x

^{2}) - integral of [-2x.sin(x

^{2})]

ln(y) = cos(x

^{2}) + 2*integral of [x.sin(x

^{2})]

*Step 5*

Using integration by parts a second time...

u = sin(x

^{2}) => du = 2x.cos(x

^{2})

dv = x => v = 1.dx

*Step 6*

Subbing back in...

ln|y| = cos(x

^{2}) + 2 ( sin(x

^{2}) - 2*integral of x.cos(x

^{2}).dx )

*Step 7*

I stop my attempt there because it just seems to eventually I get to a point where it becomes a function of itself? (ie. the x.cos(x

^{2})

What do I do next? or have I gone wrong somewhere?

EDIT --- SOME FURTHER WORK, IS THIS CORRECT?

Ok so I noticed that we get "integral of [x.cos(x

^{2})]" back in our formula, and up above in Step 2 I declared it to be = ln|y|

So I sub ln|y| into the formula and get...

ln|y| = cos(x

^{2}) + 2sin(x

^{2}) - 4*ln|y|

5*ln|y| = cos(x

^{2}) + 2sin(x

^{2})

ln|y| = (1/5)*(cos(x

^{2}) + 2sin(x

^{2}))

y = e^(above line)

Ta da? lemmy know if this correct please!

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