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Question

Often a radical change in the solution of a differential equation corresponds to a very small change in either the initial condition or the equation itself. Compare the solutions of the given initial-value problems. dydx=(y1)2,y(0)=1\frac{d y}{d x}=(y-1)^{2}, \quad y(0)=1

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Solution of initial value problem is

dydx=(y1)2y(0)=1\dfrac{dy}{dx}=\left( y-1\right)^{2} \hskip 0.5em y(0)=1

By variable separable we can write as

dy(y1)2=dx\int \dfrac{dy}{\left( y-1\right)^{2}}=\int dx

Integration will be

1y1=x+c-\dfrac{1}{y-1}=x+c

y=x+c1x+cy=\dfrac{x+c-1}{x+c}

Now, value of constant cc is by using initial value problem

1=11c1=1-\dfrac{1}{c}

Here, we can see there is no specific value of cc there is only value of c=c=\infty which satisfy above initial condition

y=1y=1

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