Question

Using separation of variable method solve dy/dx=(1-x)(1-y)​

Answer 1

`y=1-Ae^{(1)/(2)x^2-x}`

Explanation:

Given equation:

`\frac{\text{d}y}{\text{d}x}=(1-x)(1-y)`

Separate the variables by moving all the y terms to one side of the equation and all the x terms to the other side:

`\implies (1)/((1-y))\;\text{d}y=(1-x)\;\text{d}x`

Integrate both sides of the equation separately:

`\implies \displaystyle \int (1)/((1-y))\;\text{d}y= \int (1-x)\;\text{d}x`

`\implies -\ln |1-y|+C=x-(1)/(2)x^2+D`

`\implies \ln |1-y|-C=(1)/(2)x^2-x-D`

Write the two constants as one (a = -D + C):

`\implies \ln |1-y|=(1)/(2)x^2-x+a`

Take exponents of both sides:

`\implies e^(\ln |1-y|)=e^{(1)/(2)x^2-x++a}`

`\textsf{As }\; e^(\ln y)=y:`

`\implies 1-y=e^{(1)/(2)x^2-x+a}`

`\textsf{Apply exponent rule} \quad a^(b+c)=a^b \cdot a^c:`

`\implies 1-y=e^{(1)/(2)x^2-x}e^(a)`

As
`e^(a)` is just a constant, replace it with A:

`\implies 1-y=Ae^{(1)/(2)x^2-x}`

Rearrange to make y the subject:

`\implies y=1-Ae^{(1)/(2)x^2-x}`