Question

Find the particular solution of the differential equation:
√(1-x²)dy=(sin^-1x-y)dx ,it being given that x=0 when y= 0​

Answer 1

Answer:

$\sf y=(sin^{-1}x-1) +e^{-sin^{-1}x}$

Step-by-step explanation:

Given:

$\sf \sqrt{1-x^2} \:dy=(sin^{-1}x-y) \:dx$

To Find:

The particular solution of the differential equation when x = 0  and y = 0

Solution:

$\sf \sqrt{1-x^2} \:dy=(sin^{-1}x-y) \:dx$

$\sf \dfrac{dy}{dx} =\dfrac{sin^{-1}x-y}{\sqrt{1-x^2} }$

$\sf \dfrac{dy}{dx} =\dfrac{sin^{-1}x}{\sqrt{1-x^2} } -\dfrac{y}{\sqrt{1-x^2} }$

$\sf \dfrac{dy}{dx} +\dfrac{y}{\sqrt{1-x^2} }=\dfrac{sin^{-1}x}{\sqrt{1-x^2} }$

This is in the form of a linear differential equation of the type,

$\sf \dfrac{dy}{dx} + Py=Q$

where P = 1/(√1 - x²) and Q = (sin⁻¹x/√1 - x²)

Finding the integrating factor,

$\displaystyle \sf I.F=e^{\int\limits {P} \, dx}$

$\sf \implies e^{\displaystyle \sf\int\limits {\dfrac{1}{\sqrt{1-x^2} } } \, dx }$

$\sf \implies e^{sin^{-1}x}$

Now the solution of the differential equation is given by,

$\displaystyle \sf y\times (I.F)=\int\limits {Q\times (I.F)} \, dx +C$

Substitute the values,

$\displaystyle \sf y\times e^{sin^{-1}}x=\int\limits {\dfrac{sin^{-1}x}{\sqrt{1-x^2} }\times e^{sin^{-1}}x } \, dx$

Let sin⁻¹x = t

dt = 1/(√1 - x²) dx

Hence,

$\displaystyle \sf y\times e^{sin^{-1}x}=\int\limits {t\: e^t} \, dt$

Integrating by parts,

$\sf y\times e^{sin^{-1}x}=t\times e^t-e^t +C$

$\sf y\times e^{sin^{-1}x}=e^t(t-1) +C$

Give back the value of t,

$\sf y\times e^{sin^{-1}x}=e^{sin^{-1}x}(sin^{-1}x-1) +C$

Dividing by $\sf e^{sin^{-1}x}$,

$\sf y=(sin^{-1}x-1) +Ce^{-sin^{-1}x}$

Now put x = 0, y = 0,

We get,

C = 1

Therefore the particular solution of the differential equation is,

$\sf y=(sin^{-1}x-1) +e^{-sin^{-1}x}$

Answer 2

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Answer in the attachment, From my Brother