How to Solve Exact Differential Equations

Step-by-Step Guide

1. 1

   Step 1: Check if the equation is exact or not.

   In some cases, it can be trivially easy to tell if an equation is exact just from inspection.
   
   For example, ydx+xdy=0{\displaystyle y{\mathrm {d} }x+x{\mathrm {d} }y=0} is exact, because the mixed partials as stated in the introduction both equal
   1.
   
   Most other equations require that we manually take the partial derivatives and check for equality.
   
   Consider the equation below. (x+x2y2−exy3+y)dx+(23x3y−3exy2+x+2y)dy=0{\displaystyle (x+x^{2}y^{2}-e^{x}y^{3}+y){\mathrm {d} }x+\left({\frac {2}{3}}x^{3}y-3e^{x}y^{2}+x+2y\right){\mathrm {d} }y=0} Calculate partial derivatives. ∂P∂y=2x2y−3exy2+1{\displaystyle {\frac {\partial P}{\partial y}}=2x^{2}y-3e^{x}y^{2}+1} ∂Q∂x=2x2y−3exy2+1{\displaystyle {\frac {\partial Q}{\partial x}}=2x^{2}y-3e^{x}y^{2}+1} Because the partials are equal, the equation is exact.
2. 2

   Step 2: Integrate P{\displaystyle P} with respect to x{\displaystyle x} to obtain f{\displaystyle f}.

   We know that the relation P=∂f∂x{\displaystyle P={\frac {\partial f}{\partial x}}} must be true, if f{\displaystyle f} is a solution to the differential equation.
   
   However, this is a partial derivative, so doing the integration will only uncover part of the information needed.
   
   We can also choose to integrate Q{\displaystyle Q} with respect to y.{\displaystyle y.} f=∫∂f∂xdx=∫Pdx=∫(x+x2y2−exy3+y)dx=12x2+13x3y2−exy3+xy+R(y){\displaystyle {\begin{aligned}f&=\int {\frac {\partial f}{\partial x}}{\mathrm {d} }x=\int P{\mathrm {d} }x\\&=\int (x+x^{2}y^{2}-e^{x}y^{3}+y){\mathrm {d} }x\\&={\frac {1}{2}}x^{2}+{\frac {1}{3}}x^{3}y^{2}-e^{x}y^{3}+xy+R(y)\end{aligned}}} The constant of integration R(y){\displaystyle R(y)} represents this incomplete information, in that any arbitrary function of y{\displaystyle y} only will satisfy P.{\displaystyle P.} If we chose to integrate Q,{\displaystyle Q,} our constant of integration will be with respect to x{\displaystyle x} only. , Likewise, the condition ∂f∂y=Q{\displaystyle {\frac {\partial f}{\partial y}}=Q} must hold as well.
   
   What we find is that the terms in Q{\displaystyle Q} that are purely a function of y{\displaystyle y} equal dRdy.{\displaystyle {\frac {{\mathrm {d} }R}{{\mathrm {d} }y}}.} If we did it the other way, then we would take the derivative with respect to x{\displaystyle x} and find that the terms in P{\displaystyle P} that are purely a function of x{\displaystyle x} must satisfy dRdx.{\displaystyle {\frac {{\mathrm {d} }R}{{\mathrm {d} }x}}.} ∂f∂y=23x3y2−3exy2+dRdy=Q{\displaystyle {\frac {\partial f}{\partial y}}={\frac {2}{3}}x^{3}y^{2}-3e^{x}y^{2}+{\frac {{\mathrm {d} }R}{{\mathrm {d} }y}}=Q} By inspection, we can see that dRdy=2y.{\displaystyle {\frac {{\mathrm {d} }R}{{\mathrm {d} }y}}=2y.} , With practice, you can jump straight to inspecting Q{\displaystyle Q} for any pure y{\displaystyle y} terms (or P{\displaystyle P} for any pure x{\displaystyle x} terms, if we did it the other way) and integrating them.
   
   R=y2{\displaystyle R=y^{2}} , The solution contains an arbitrary constant C,{\displaystyle C,} as usual.
   
   We also see that this function is defined implicitly.
   
   In general, this will be the case, as it can be difficult to express the solution in explicit form. 12x2+13x3y2−exy3+xy+y2=C{\displaystyle {\frac {1}{2}}x^{2}+{\frac {1}{3}}x^{3}y^{2}-e^{x}y^{3}+xy+y^{2}=C}
3. 3

   Step 3: Take the derivative of f{\displaystyle f} with respect to y{\displaystyle y}.

4. 4

   Step 4: Find the constant of integration R{\displaystyle R}.

5. 5

   Step 5: Arrive at the solution.

Detailed Guide

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