The family of solutions to the differential equation y ′ = −4xy3 is y = 1 √ C + 4x2 . Find the solution that satisfies the initial condition y(−2) = 4.

1. y = 4 p 1 + 16(x2 − 4)
2. y = 1 p 1 + 64(x 2 − 4)
3. y = 4 p 1 + 4(x2 − 4)
4. y = 4 p 1 + 64(x 2 − 4)
5. y = 1 p −4(x2 − 4) + 1

[tex]y(x)=\frac{1}{\sqrt{C+4x^2}}\\y(-2)=\frac{1}{\sqrt{C+4(-2)^2}}\\4=\frac{1}{\sqrt{C+4(4)}}\\4=\frac{1}{\sqrt{C+16}}\\16=\frac{1}{C+16}\\C+16=\frac{1}{16}\\C=\frac{1}{16}-16[/tex]

So the solution is given as

[tex]y(x)=\frac{1}{\sqrt{C+4x^2}}\\y(x)=\frac{1}{\sqrt{\frac{1}{16}-16+4x^2}}[/tex]

Simplifying the equation as

[tex]y(x)=\frac{1}{\sqrt{\frac{1}{16}-16+4x^2}}\\y(x)=\frac{1}{\sqrt{\frac{1-256+64x^2}{16}}}\\y(x)=\frac{\sqrt{16}}{\sqrt{{1-256+64x^2}}}\\y(x)=\frac{4}{\sqrt{{1+64(x^2-4)}}}[/tex]

So the correct option is 4

Hanneda Hanneda · Answer 2 · 2020-02-24T15:48:32+01:00

Hanneda Hanneda

24-02-2020

Answer:

All the answers are incorrect

Step-by-step explanation:

We substituted the indicated value of y in to the solution y(x) inorder to solve for C as follows:

[\tex]y(-2)=1\frac{1}{\sqrt{C+4(-2)^2}}=4[\tex]

[\tex]1\frac{1}{\sqrt{C+16}}=4[\tex]

Squaring both sides, we have

[\tex]1\frac{1}{C+16}=16[\tex]

Therefore, [\tex]C =\frac{1}{16}-16[\tex]

The solution is now

[\tex]y(-2)=1\frac{1}{\sqrt{\frac{1}{16}-16+4x^2}}=4[\tex]

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