Question

(1-x2) dy/dx +xy=y3.sin-1x​

Answer 1

Explanation:

this problem is from differential equations

Answer 2

Answer:

The solution of the given differential equation will be $(1-x^{2} /y^{2} ) = -2(x.sin^{-1}x+\sqrt{1-x^{2} } ) +c$  

Explanation:

• The given differential equation is: (1-x²)dy/dx + xy = y³$sin^{-1} x$
• Dividing both sides with (1-x²): dy/dx + xy/(1-x²) = y³$sin^{-1} x$/(1-x²)

• In Bernoulli's equation: dy/dx + y.P(x) = Q.$y^{n}$ [where P(x) and Q is a function of x or constant, n is a real number n≠0,n≠1 ]

• We have transformed the given equation into Bernoulli's equation. Where P(x) = x/(1-x²) and Q = $sin^{-1} x$/(1-x²) , n=3.
• Dividing both sides with y³ : 1/y³dy/dx + x/(1-x²)y² = $sin^{-1} x$/(1-x²)
• Let u = $y^{-2}$ , Differentiate both sides w.r.t x

                                    du/dx = (-2$y^{-3}$)dy/dx

                               ⇒  (-1/2)du/dx = 1/y³(dy/dx)

• The equation becomes: (-1/2)du/dx + (x/1-x²)u = $sin^{-1}x$/(1-x²)
• Multiplying both sides with -2: du/dx -2x/(1-x²)u = -2$sin^{-1} x$/(1-x²)
• By comparing the equation to Bernoulli's equation. Where P(x) = -2x/(1-x²) and Q = -2 $sin^{-1} x$/(1-x²)
• Integration factor(I.F): $e^{\int\limits^.{P} \, dx }$.
• Putting the value of P in I.F : $e^{\int\limits^.{-2x/(1-x^{2} )} \, dx }$
• After solving this I.F will be: (1-x²)

• The general solution of the given equation:

                        $u(1-x^{2} )=$ $\int\limits {-2sin^{-1} }x.(1-x^{2} )/(1-x^{2} ) \, dx +c$    

                      ⇒  $u(1-x^{2} )=-2\int\limits{sin^{-1} }x \, dx +c$

                      ⇒ $u(1-x^{2} ) = -2(x.sin^{-1}x+\sqrt{1-x^{2} } ) +c$

                      ⇒ $(1-x^{2} /y^{2} ) = -2(x.sin^{-1}x+\sqrt{1-x^{2} } ) +c$......(Ans)

To learn more about differential equations:

https://brainly.com/question/1164377

To learn more about Bernoulli's equation:

https://brainly.in/question/7079240

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