Differential equation
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Is this meant to read:

dy/dx=[(x-1)(x+2)³-2y(x+5)]/(x²+y-2) or

dy/dx=(x-1)(x+2)³-[2y(x+5)/(x²+y-2)]?

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1 Answer

It’s not clear whether the divisor is applied to both terms on the right-hand side or just the second term. If we assume the former then we have a DE of the form:

dy/dx=(R(x)-P(x,y))/Q(x,y)

which can be written:

P(x,y)+Q(x,y)dy/dx=R(x).

If this DE is exact, then ∂P/∂y=∂Q/∂x where P(x,y)=2y(x+5), so ∂P/∂y=2(x+5), and Q(x,y)=x²+y-2, so ∂Q/∂x=2x, and the partial derivatives are unequal.

We may be able to find an integrating factor I(x) or J(y) which makes the partial derivatives equal. If such is the case, we have ∂(IP)/∂y=∂(IQ)/∂x or ∂(JP)/∂y=∂(JQ)/∂x.

∂(IP)/∂y=∂(IQ)/∂x is the same as I∂P/∂y=QdI/dx+I∂Q/∂x⇒QdI/dx=I(∂P/∂y-∂Q/∂x), ∫dI/I=∫(1/Q)(∂P/∂y-∂Q/∂x)dx, and ln(I)=∫(1/Q)(∂P/∂y-∂Q/∂x)​dx, I=e^∫(1/Q)(∂P/∂y-∂Q/∂x)​dx. [The partial derivative of I wrt x is the same as dI/dx, and similarly the PD of J wrt y is dJ/dy. I is treated as a constant when finding the PD wrt y, and J is treated as a constant when finding the PD wrt x.]

Similarly, J=e^∫(1/P)(∂Q/∂x-∂P/∂y)​dy.

∂P/∂y-∂Q/∂x=2x+10-2x=10 in this case, substituting for P and Q.

I must be a function of x only and J must be a function of y only. However, because neither 10/Q or -10/P gives a reducible expression in x or y alone, an integrating factor cannot be found using this method.

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