find the solution of
(x2-y2)dx + 3xydy =0
Please help me to find the right answer,I sovled many ways but I don't get the right one with answer key.
Thank you so much in advance
To find the solution of the given differential equation (x² - y²)dx + 3xydy = 0, we can use the method of exact differential equations.
Step 1: Check for Exactness
To determine if the equation is exact, we need to check if the partial derivatives of the coefficients with respect to y (✶) and x (✷) are equal.
In our case:
✶ ∂/∂y (x² - y²) = -2y
✷ ∂/∂x (3xy) = 3y
Since -2y is not equal to 3y, the equation is not exact. Therefore, we move on to the next step.
Step 2: Multiply through by an Integrating Factor
To make the equation exact, we multiply through by an integrating factor, which is the reciprocal of the function multiplying dy. In our case, the function is 3xy.
The integrating factor (IF) is given by:
IF = e^(∫(✶-✷)/x) dx
Substituting the given values:
IF = e^(∫(-2y - 3y)/x) dx
= e^(∫-5y/x) dx
= e^(-5∫y/x) dx
= e^(-5ln|x|) dx
= 1/x^5
Multiply the entire equation by 1/x^5:
(x² - y²)/x^5 dx + (3x/y)/x^5 dy = 0
Step 3: Apply Exact Differential Equation Technique
Now that we have an exact equation, we can find the solution by integrating both sides with respect to x or y and adding a constant, if necessary.
Integrating the given equation:
∫(x² - y²)/x^5 dx + ∫(3x/y)/x^5 dy = 0
Simplifying the integrals:
∫(x^-3 - y²/x^5) dx + ∫(3/x^4) dy = 0
(-1/2)x^-2 - (-1/4)y^4/x^4 + (3/x^4)y + C = 0
Rearranging the terms:
(-1/2)x^-2 - (1/4)(y^4/x^4) + (3/x^4)y + C = 0
This is the most simplified form of the solution to the given differential equation (x² - y²)dx + 3xydy = 0.
If you were unable to obtain the same result using a different method, it is possible that there was an error in the calculation or the answer key may contain a mistake. Double-check your steps and calculations to ensure accuracy.