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Find the solution to the exact differential equation

(xy²+y²+y³) dx+(3y²x+2yx+x²y) dy=0, y(x)=? ? ?

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Final answer:

To solve the given exact differential equation, we first check if it is exact. Then, we integrate the equation to find the solution function. The solution is given by the equation: x(y^2x) + y^3(x) + y^2(y) + xy^2 + (3/2)(y^2x^2) = C, where C is the constant of integration.

Step-by-step explanation:

To find the solution to the given exact differential equation, we need to check if it is exact and find the integrating factor if necessary. Let's check if it is exact:

Step 1: Calculate the partial derivatives of the function with respect to x (fx) and y (fy):

fx = y^2 + y^3

fy = 3yx^2 + 2xy + y^2

Step 2: Check if the equation fx dx + fy dy = 0 satisfies the condition fy = ∂f/∂y, where f(x, y) is the solution function of the differential equation:

fy = ∂f/∂y = 2xy + y^2

Since fy = ∂f/∂y, the equation is exact. Therefore, we can find the solution function by integrating the equation:

∫(xy^2 + y^3)dx + ∫(3yx^2 + 2xy + y^2)dy = 0

To solve this, integrate the first term with respect to x and the second term with respect to y:

∫xy^2 dx + ∫y^3 dx + ∫3yx^2 dy + ∫2xy dy + ∫y^2 dy = 0

x∫y^2 dx + y^3∫dx + y^2∫dy + 2∫xy dy + 3∫yx^2 dy = 0

x(y^2x) + y^3(x) + y^2(y) + 2(xy^2)/2 + 3(y^2x^2)/2 = C

where C is the constant of integration.

User Simone Gianni
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