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Solution of the differential equation is...

Solution of the differential equation is
`(1-x^(2))(dy//dx) +2xy =x sqrt("") (1-x^(2))` Then
`ysqrt("") (1-x^(2)) =c (1-x^(2))` . True or False

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The correct Answer is:
To solve the differential equation \((1-x^2)\frac{dy}{dx} + 2xy = x\sqrt{1-x^2}\), we will follow these steps: ### Step 1: Rearranging the Equation We start by rearranging the given differential equation to isolate \(\frac{dy}{dx}\): \[ (1-x^2)\frac{dy}{dx} = x\sqrt{1-x^2} - 2xy \] Now, we can express \(\frac{dy}{dx}\) as: \[ \frac{dy}{dx} = \frac{x\sqrt{1-x^2} - 2xy}{1-x^2} \] ### Step 2: Separating Variables Next, we can separate variables. We rewrite the equation: \[ \frac{dy}{\sqrt{1-x^2}} = \frac{x - 2y}{1-x^2}dx \] This allows us to integrate both sides separately. ### Step 3: Integrating Both Sides We integrate the left side with respect to \(y\) and the right side with respect to \(x\): \[ \int dy = \int \frac{x - 2y}{1-x^2}dx \] The left side integrates to \(y\), and we need to evaluate the right side. ### Step 4: Solving the Right Side Integral The right side can be split into two parts: \[ \int \frac{x}{1-x^2}dx - 2\int \frac{y}{1-x^2}dx \] The first integral can be solved using substitution, and the second integral can be integrated directly. ### Step 5: Finding the General Solution After integrating, we combine the results and include a constant of integration \(C\): \[ y\sqrt{1-x^2} = C(1-x^2) \] ### Conclusion Thus, the solution of the differential equation is: \[ y\sqrt{1-x^2} = C(1-x^2) \] ### Final Statement The statement given in the question is **True**.
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Solution of the differential equation (1+x^(2)) (dy//dx) +2xy = cos x is y(1+x^(2)) = sin x +c .

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Knowledge Check

  • Solution of the differential equatio (1-x^(2))(dy)/(dx)+2xy=xsqrt(1-x^(2)) is

    A
    `y-sqrt(1-x^(2))=c(1-x^(2))`
    B
    `y+sqrt(1-x^(2))=c(1-x^(2))`
    C
    `y-2sqrt(1-x^(2))=c(1-x^(2))`
    D
    `y+2sqrt(1-x^(2))=c(1-x^(2))`
  • Solution of the differential equation (1+x)(dy)/(dx)-xy=1-x is

    A
    `xy+y+x=ce^(x)`
    B
    `xy+y-x=ce^(x)`
    C
    `xy+y+x=ce^(-x)`
    D
    `xy+y-x=ce^(-x)`
  • Solution of the differential equation (1+x^(2)) dy + 2xy dx = cot x dx is

    A
    ` y = log |sin x | (1+x^(2))+C(1+x^(2))^(-1)`
    B
    ` y = log |sin x|(1+x^(2))^(-1)+C(1+x^(2))`
    C
    ` y = log |sin x | (1+x^(2))+C(1+x)`
    D
    ` y = log |sin x| (1+x^(2))^(-1)`
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