# `dy/dx = (1-2x) / (4x-x^2)` Solve the differential equation

`dy/dx = (1-2x) / (4x-x^2)`

This differential equation is separable since it can be written in the form

• `N(y)dy =M(x)dx`

Bringing together same variables on one side, the equation becomes

`dy=(1-2x)/(4x-x^2)dx`

Taking the integral of both sides, it turns into

`int dy = int(1-2x)/(4x-x^2)dx`

`y+C_1 = int(1-2x)/(4x-x^2)dx`

`y+C_1= int (1-2x)/(x(4-x))dx`

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`dy/dx = (1-2x) / (4x-x^2)`

This differential equation is separable since it can be written in the form

• `N(y)dy =M(x)dx`

Bringing together same variables on one side, the equation becomes

`dy=(1-2x)/(4x-x^2)dx`

Taking the integral of both sides, it turns into

`int dy = int(1-2x)/(4x-x^2)dx`

`y+C_1 = int(1-2x)/(4x-x^2)dx`

`y+C_1= int (1-2x)/(x(4-x))dx`

To take the integral of right side, apply partial fraction decomposition.

• `(1-2x)/(x(4-x)) = A/x + B/(4-x)`
• `1-2x = A(4-x)+Bx`

Let x=0.

`1-2(0) = A(4-0)+B(0)`

`1=4A`

`1/4=A`

Let x=4.

`1-2(4)=A(4-4)+B(4)`

`-7=4B`

`-7/4=B`

• `1/(2x-x^2) = (1/4)/x + (-7/4)/(4-x)`
• `1/(2x-x^2) = 1/(4x) + (-7)/(4(4-x))`
• `1/(2x-x^2) = 1/(4x) + (-7)/(-4(x-4))`
• `1/(2x-x^2) = 1/(4x) + (7)/(4(x-4))`

So the integrand at the right side decomposes to

`y + C_1 = int (1/(4x) + 7/(4(x-4)))dx`

Then, apply the formula `int 1/u du = ln|u| + C` .

`y + C_1 = 1/4ln|x| + 7/4ln|x-4|+C_2`

Isolating the y, the equation becomes

`y= 1/4ln|x| + 7/4ln|x-4|+C_2-C_1`

Since C1 and C2 represent any number, it can be expressed as a single constant C.

`y = 1/4ln|x| + 7/4ln|x-4|+C`

Therefore, the general solution of the given differential equation is `y = 1/4ln|x| + 7/4ln|x-4|+C` .

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