Question

Solve differential equationdy/dx+xy=xy^2

First order differential equations
ANSWERED
asked 2021-01-13

Solve differential equation \(\frac{dy}{dx}+xy=xy^2\)

Answers (1)

2021-01-14

\(\frac{dy}{dx}+xy=xy^2\)
\(\Rightarrow \frac{dy}{dx}= xy^2-xy\)
\(\Rightarrow \frac{dy}{dx}= xy(y-1)\)
\(\Rightarrow \frac{dy}{y(y-1)}= xdx\)
\(\frac{1}{y(y-1)}= \frac{A}{y}+\frac{B}{y-1}\)
\(\Rightarrow1= A(y-1)+By\)
\(\Rightarrow1= (A+B)y-A\)
Comparing the coefficient of y and constant term on both side, we get
\(A+B=0,\ -A=1\Rightarrow A= -1,\ B=1\)
So, we write
\(\frac{1}{y(y−1)}= -\frac{1}{y}+\frac{1}{(y−1)}\)
So, we have
\((-\frac{1}{y}+\frac{1}{(y−1)}) dy= xdx\)
Now, we will integrate this equation on both the side
\(\int (-\frac{1}{y}+\frac{1}{(y-1)})dy= \int xdx\) \(\Rightarrow -\ln(y)+\ln(y-1)= x^{\frac{2}{2}+C}\)
\(\Rightarrow \frac{y-1}{y}= x^{\frac{2}{2}+C}\)
\(\Rightarrow \frac{y-1}{y}= e^{x^{\frac{2}{2}+C}}\)
\(\Rightarrow \frac{y-1}{y}= e^{x^{\frac{2}{2}}}e^{C}\)
\(\Rightarrow \frac{y-1}{y}= C_{1} e^{x^{\frac{2}{2}}}\), \(C_1=e^C\)
where C and \(C_1\) are arbitrary constant
\(\Rightarrow y-1= C_{1} e^{\frac{x^{\frac{2}{2}}}{y}}\)
\(\Rightarrow y= C_{1} e^{x^{\frac{2}{2}}y}+1\)
\(\Rightarrow y-C e^{x^{\frac{2}{2}}}y= 1\)
\(\Rightarrow y(1-C_{1} e ^{x^{\frac{2}{2}}})=1\)
\(\Rightarrow y= \frac{1}{1-C_{1} e^{x^{\frac{2}{2}}}}\)

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