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Free_Kalibri [48]
2 years ago
13

If the graph of y=f(x) passes through the point (0,1), and dy/dx=xsin(x^2)/y, then f(x)= ?

Mathematics
1 answer:
Zigmanuir [339]2 years ago
4 0

The differential equation

d<em>y</em>/d<em>x</em> = <em>x</em> sin(<em>x</em> ²) / <em>y</em>

is separable as

<em>y</em> d<em>y</em> = <em>x</em> sin(<em>x</em> ²) d<em>x</em>

Integrate both sides:

∫ <em>y</em> d<em>y</em> = ∫ <em>x</em> sin(<em>x</em> ²) d<em>x</em>

∫ <em>y</em> d<em>y</em> = 1/2 ∫ 2<em>x</em> sin(<em>x</em> ²) d<em>x</em>

∫ <em>y</em> d<em>y</em> = 1/2 ∫ sin(<em>x</em> ²) d(<em>x</em> ²)

1/2 <em>y</em> ² = -1/2 cos(<em>x</em> ²) + <em>C</em>

Solve for <em>y</em> implicitly:

<em>y</em> ² = -cos(<em>x</em> ²) + <em>C</em>

Given that <em>y</em> = 1 when <em>x</em> = 0, we get

1² = -cos(0²) + <em>C</em>

1 = -cos(0) + <em>C</em>

1 = -1 + <em>C</em>

<em>C</em> = 2

Then the particular solution to the DE is

<em>y</em> ² = 2 - cos(<em>x</em> ²)

Solving explicitly for <em>y</em> would give two solutions,

<em>y</em> = ± √(2 - cos(<em>x</em> ²))

but only the one with the positive square root satisfies the initial condition:

√(2 - cos(0²)) = √1 = 1

-√(2 - cos(0²)) = -√1 = -1 ≠ 1

So the unique solution to this DE is

<em>y</em> = √(2 - cos(<em>x</em> ²))

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