## Related questions with answers

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. $\frac{d y}{d x}=(y-1)^{2}, \quad y(0)=1$

Solution

VerifiedStep 1

1 of 2Solution of initial value problem is

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

By variable separable we can write as

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

Integration will be

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

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

Now, value of constant $c$ is by using initial value problem

$1=1-\dfrac{1}{c}$

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

$y=1$

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