Step-by-step Solution

Solve the differential equation $\left(2x-1\right)dx+\left(3y+7\right)dy=0$

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Step-by-step explanation

Problem to solve:

$\left(2x-1\right)\cdot dx+\left(3y+7\right)\cdot dy=0$

Learn how to solve differential equations problems step by step online.

$\left(2x-1\right)dx+\left(3y+7\right)dy=0$

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Learn how to solve differential equations problems step by step online. Solve the differential equation (2x-1)dx+(3y+7)dy=0. The differential equation \left(2x-1\right)dx+\left(3y+7\right)dy=0 is exact, since it is written in the standard form M(x,y)dx+N(x,y)dy=0, where M(x,y) and N(x,y) are the partial derivatives of a two-variable function f(x,y) and they satisfy the test for exactness: \displaystyle\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}. In other words, their second partial derivatives are equal. The general solution of the equation is f(x,y)=C. Using the test for exactness, we check that the differential equation is exact. Integrate M(x,y) with respect to x to get. Now take the partial derivative of x^2-x with respect to y to get.

Final Answer

$x^2-x+\frac{3}{2}y^2+7y=C_0$

Problem Analysis

$\left(2x-1\right)\cdot dx+\left(3y+7\right)\cdot dy=0$

Time to solve it:

~ 0.04 seconds

Related topics:

Differential equations