Answer:
[tex]f(x,y)=2x^2+4y^2+2xy=C_1\\\\Where\\\\y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )[/tex]
Step-by-step explanation:
Let:
[tex]M(x,y)=4x+2y\\\\and\\\\N(x,y)=2x+8y[/tex]
This is and exact equation, because:
[tex]\frac{\partial M(x,y)}{\partial y} =2=\frac{\partial N}{\partial x}[/tex]
So, define f(x,y) such that:
[tex]\frac{\partial f(x,y)}{\partial x} =M(x,y)\\\\and\\\\\frac{\partial f(x,y)}{\partial y} =N(x,y)[/tex]
The solution will be given by:
[tex]f(x,y)=C_1[/tex]
Where C1 is an arbitrary constant
Integrate [tex]\frac{\partial f(x,y)}{\partial x}[/tex] with respect to x in order to find f(x,y):
[tex]f(x,y)=\int\ {4x+2y} \, dx =2x^2+2xy+g(y)[/tex]
Where g(y) is an arbitrary function of y.
Differentiate f(x,y) with respect to y in order to find g(y):
[tex]\frac{\partial f(x,y)}{\partial y} =2x+\frac{d g(y)}{dy}[/tex]
Substitute into [tex]\frac{\partial f(x,y)}{\partial y} =N(x,y)[/tex]
[tex]2x+\frac{dg(y)}{dy} =2x+8y\\\\Solve\hspace{3}for\hspace{3}\frac{dg(y)}{dy}\\\\\frac{dg(y)}{dy}=8y[/tex]
Integrate [tex]\frac{dg(y)}{dy}[/tex] with respect to y:
[tex]g(y)=\int\ {8y} \, dy =4y^2[/tex]
Substitute g(y) into f(x,y):
[tex]f(x,y)=2x^2+4y^2+2xy[/tex]
The solution is f(x,y)=C1
[tex]f(x,y)=2x^2+4y^2+2xy=C_1[/tex]
Solving y using quadratic formula:
[tex]y(x)=\frac{1}{4} (-x\pm \sqrt{-7x^2+C_1} )[/tex]
