Solve a differential equation:&nbsp;dy/dx=x2y2+x2-y2-1

Here is a systematic way: assuming your right-hand side, call it f(x,y)=x2y2+x2−y2−1, actually is separable as f(x,y)=X(x)Y(y), then this will work for any x0 and y0. So arbitrarily choosing (x0,y0)=(0,0) for simplicity, we haveX(x)∞X(x)Y(y0)X(x0)Y(y0)=f(x,y0)f(x0,y0)=x2−1−1=1−x2.Now with X(x)=1−x2, we get the other factorY(y)∞X(x0)Y(y)=f(x0,y)=−y2−1and lo and behold, X(x)=1−x2 and Y(y)=−y2−1 give f(x,y).In taking X(x)=1−x2, I threw out the constant of proportionality 1X(x0), and in taking Y(y)=−y2−1, I threw out (x0). But it's OK since they would have cancelled out anyway. I discovered this shortcut after getting tired of guessing on this type of problem.

Check the solution to "Solve a differential equation:dy/dx=x2y2+x2-y2-1"

Arthur Pratt

Answered question

2022-01-20

Solve a differential equation: $d y / d x = x^{2} y^{2} + x^{2} - y^{2} - 1$

Answer & Explanation

Cleveland Walters

Beginner2022-01-20Added 40 answers

Here is a systematic way: assuming your right-hand side, call it

f (x, y) = x^{2} y^{2} + x^{2} - y^{2} - 1

, actually is separable as

f (x, y) = X (x) Y (y)

, then this will work for any

x_{0} and y_{0}

. So arbitrarily choosing

(x_{0}, y_{0}) = (0, 0)

for simplicity, we have

X (x) \infty \frac{X (x) Y (y_{0})}{X (x_{0}) Y (y_{0})} = \frac{f (x, y_{0})}{f (x_{0}, y_{0})} = \frac{x^{2} - 1}{- 1} = 1 - x^{2}

.
Now with

X (x) = 1 - x^{2}

, we get the other factor

Y (y) \infty X (x_{0}) Y (y) = f (x_{0}, y) = - y^{2} - 1

and lo and behold,

X (x) = 1 - x^{2} and Y (y) = - y^{2} - 1

give f(x,y).
In taking

X (x) = 1 - x^{2}

, I threw out the constant of proportionality

\frac{1}{X} (x_{0})

, and in taking

Y (y) = - y^{2} - 1

, I threw out

(x_{0})

. But it's OK since they would have cancelled out anyway. I discovered this shortcut after getting tired of guessing on this type of problem.

Rita Miller

Beginner2022-01-21Added 28 answers

You can factor the right side as

(y^{2} + 1) (x^{2} - 1)

Do you have a similar question?

Recalculate according to your conditions!

New Questions in Differential Equations

Didn't find what you were looking for?