Question

a. Solve the differential equation y'=4x(sqrt(1-(y^2)).
b. Explain why the initial value problem y'=4x(sqrt(1-(y^2)) with y(0) = 4 does not have a solution.

Answer 1

\[y'=4x \sqrt{1-y ^{2}} \]

Answer 2

\[ \frac{ dy }{ dx }=4x \sqrt{1-y ^{2}}\]

\[\frac{ 1 }{ \sqrt{1-y ^{2}} } dy= 4x dx\]

\[\int\limits \frac{ 1 }{ \sqrt{1-y ^{2}} } dy= \int\limits 4x dx\]

\[\arcsin(x)+C=2x ^{2}+C\]

\[\arcsin(x)=2x ^{2}+C\]

Answer 3

I understand that we can not plug a y value in because there is simply none. But what I want to know is how to solve this equation

Answer 4

The left hand side should be \(\arcsin \color{red}y\) not \(\arcsin x\). You were integrating with respect to \(y\) to begin with.

Answer 5

\[\arcsin(y)=2x ^{2}+C\]

Answer 6

alright let me see what i can get from there

Answer 7

In any case, \(\arcsin y\) is not defined for \(y=4\) because \(4\) is not in the domain of the inverse sine function. Hence no solution can be obtained.

Answer 8

@SithsAndGiggles I see what you mean. But in part a the question says "Solve the differential equation y'=4x(sqrt(1-(y^2))." so how would i "solve" the equation; woulod i simply state the furtherest that i can go is arcsin(y)=2x^2+C?

Answer 9

becuase of said situation mentioned above? @SithsAndGiggles

Answer 10

You can still solve for \(y\) to get \(y=\sin(2x^2+C)\) but you also have to provide the interval over which this solution is valid. Namely, \(\left|2x^2+C\right|\le\dfrac{\pi}{2}\). This is because \(f(x)=\arcsin x\) has a domain of \([-1,1]\) and range \(-\dfrac{\pi}{2}\le f(x)\le\dfrac{\pi}{2}\).

Answer 11

@SithsAndGiggles okay so given the interval what's the solution and how do you get it?

Answer 12

You already have the solution. You can leave it as \(\arcsin y=2x^2+C\). This answers (a). No solution exists for the initial value \(y(0)=4\) because the inverse sine is not defined for values of \(y\) less than \(-1\) and greater than \(1\).

Answer 13

@SithsAndGiggles okay thanks :)