Solve the following differential equation by finding an appropriate integrating factor -xdx+(3y-x^2y)dy=0

Question

 Solve the following differential equation by finding an appropriate integrating factor

-xdx+(3y-x^2y)dy=0

Constant =??

I have calculated 1/2y^2+1/2ln|3-x^2|= c & y^2+ln|3-x^2|

both however are incorrect can someone please help me?

Answer 1

This is of the form M(x,y)dx + N(x,y)dy = 0

But it is not exact because M_y ≠ N_x, so we need to use the formula

μ(x) = (M_y - N_x)/N = (0-2x)/(3y-x²)y = -2x/(3-x²)

Integrating factor, If = e^∫μ(x)dx = e^{ln(3-x^2)} = 3-x²

Multiplying through by (3-x²)

(3-x²)(-x) + (3-x²)y(3-x²)dy/dx = 0

separating variables

∫ydy = ∫(x/(3-x²)dx = -ln|3-x^2|/2 + C₁

y²/2 = C₁-ln|3-x²)|2

y = ±√(C₂-ln|3-x²|)

Answer 2

I AGREE WITH YOU!!!!

-x dx + (3y - x^2 y ) dy = 0

Factors out y in the second term...

-x dx + y (3-x^2) dy = 0

divides by (3-x^2)...

-x/(3-x^2) + y dy = 0

moves first term to right side....

y dy = x dx/(3-x^2)

For integration, let U=3-x^2

dU = -2x dx

(-1/2) dU = x dx

Integrates both sides:

integral [ y dy] = (-1/2) integral [ dU/U]

(1/2)y^2 = (-1/2) ln |U| + c = (-1/2) ln | 3 - x^2 | + c

Next, the 1/2 cancels....

y^2 = - [ln | 3 - x^2| + c] = c - ln |3 - x^2 |

y = sqrt[ c - ln |3 - x^2 |]

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Checks by differentiation:

dy/dx = (1/2) [ c - ln|3-x^2|]^(-1/2) *(-2x)/(3-x^2)

= -x/ {[ c - ln|3-x^2|]^(1/2) * (3-x^2)}

= -x/ {sqrt [ c - ln|3-x^2|] * (3-x^2)}

= -x / {y * (3-x^2)}

Cross multiplies:

y * (3-x^2)*dy = -x dx

which is the original differential equation

once the term on the right is moved to the left...

So this answer is correct despite what the computer

says. Perhaps it is expecting the answer in a different

format. Are you inputting the square root correctly?