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that's why i multiply the whole equation with the integrating factor to get the exact equation , however , i didnt get it . Is my working totally wrong ? or which part is wrong ?HallsofIvy said:The formula you are using to find the integrating factor only applies to linear equations. This is not a linear equation!
What integrating factor are you talking about? Your work up to the point where you say that the equation is not exact, but the work after that, I don't follow.goldfish9776 said:that's why i multiply the whole equation with the integrating factor to get the exact equation , however , i didnt get it . Is my working totally wrong ? or which part is wrong ?
That's what the OP is trying to do. If the R(⋅) expressions had turned out to be functions of just x or just y, integrating them would yield the integrating factor. (Of course, it would have been nice if the OP had written the post properly and explained this instead of making us guess at what he/she was doing.)Mark44 said:What integrating factor are you talking about? Your work up to the point where you say that the equation is not exact, but the work after that, I don't follow.
Since the DE is not exact, you might look for an integrating factor that is a function of x alone. If that doesn't work, you can see if there is an integrating factor of y alone. Finally, if that doesn't lead to an exact equation, you can try an integrating factor that is a function of both x and y.
do u mean R(x , y) , how to do it ? i have never studied this beforeMark44 said:What integrating factor are you talking about? Your work up to the point where you say that the equation is not exact, but the work after that, I don't follow.
Since the DE is not exact, you might look for an integrating factor that is a function of x alone. If that doesn't work, you can see if there is an integrating factor of y alone. Finally, if that doesn't lead to an exact equation, you can try an integrating factor that is a function of both x and y.
i still didnt get the idea, can you explain further?vela said:You might want to take a look at http://mathworld.wolfram.com/ExactFirst-OrderOrdinaryDifferentialEquation.html to get ideas. I don't think any of the specific cases they looked at works for this particular problem.
can you be fore specific , telling me how should i start with ?vela said:Not really. All I'd be doing is repeating what's on that page, which is pretty clear. Maybe someone else can chime in.
An integrating factor is a function used in solving differential equations by multiplying both sides of the equation. It helps to simplify complex equations and make them easier to solve.
An integrating factor should be used when solving first-order linear differential equations that involve variables that are not separable. It can also be used to solve non-linear equations, but this is more complex and requires more advanced techniques.
The integrating factor can be determined by finding the coefficient of the highest derivative in the differential equation. This coefficient is then used as the base for the integrating factor, which is usually expressed as e to the power of an integral. The integral is then solved to find the exact value of the integrating factor.
No, an integrating factor can only be used in certain types of differential equations, such as first-order linear equations. It cannot be used in equations that are already separable or do not involve variables that can be separated.
The integrating factor helps to simplify the equation and make it easier to solve. It can also change the form of the solution, making it easier to identify and solve. Additionally, it allows for the use of more straightforward techniques, such as integration by parts, to solve more complex equations.