Let the solution curve of the differential equation passes through the origin. Then is equal to.

Question

Let the solution curve $y = f (x)$ of the differential equation $\frac{d y}{d x} + \frac{x y}{x^{2} - 1} = \frac{15 x^{4} + 4 x^{3} + 4 x + 1}{\sqrt{1 - x^{2}}}, x \in (- 1, 1)$ passes through the origin. Then $\int_{- 1 / 2}^{1 / 2} \frac{f (x) d x}{x^{2} + 2}$ is equal to.

Smart Achievers · Accepted Answer

(2)

$I . F . = e^{\int \frac{x d x}{x^{2} - 1}} = \sqrt{1 - x^{2}}$

$y = f (x) = \frac{3 x^{5} + x^{4} + 2 x^{2} + x}{\sqrt{1 - x^{2}}}$

$= \int_{- 1 / 2}^{1 / 2} \frac{f (x)}{(x^{2} + 2)} d x = \frac{π}{6} - \frac{1}{4}$

Solution

Q.
Let the solution curve $y = f (x)$ of the differential equation $\frac{d y}{d x} + \frac{x y}{x^{2} - 1} = \frac{15 x^{4} + 4 x^{3} + 4 x + 1}{\sqrt{1 - x^{2}}}, x \in (- 1, 1)$ passes through the origin. Then $\int_{- 1 / 2}^{1 / 2} \frac{f (x) d x}{x^{2} + 2}$ is equal to.

1 $\frac{π}{6}$

2 $\frac{π}{6} - \frac{\sqrt{3}}{4}$

3 $\frac{π}{3} - \frac{\sqrt{3}}{4}$

4 $\frac{π}{3} - \frac{1}{4}$

Ans.
(2)

$I . F . = e^{\int \frac{x d x}{x^{2} - 1}} = \sqrt{1 - x^{2}}$

$y = f (x) = \frac{3 x^{5} + x^{4} + 2 x^{2} + x}{\sqrt{1 - x^{2}}}$

$= \int_{- 1 / 2}^{1 / 2} \frac{f (x)}{(x^{2} + 2)} d x = \frac{π}{6} - \frac{1}{4}$

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Solution

Q. Let the solution curve y=f(x) of the differential equation dydx+xyx2-1=15x4+4x3+4x+11-x2, x∈(-1,1) passes through the origin. Then ∫-1/21/2f(x) dxx2+2 is equal to.

1 π6

2 π6-34

3 π3-34

4 π3-14

Ans. (2) I.F.=e∫x dxx2-1=1-x2 y=f(x)=3x5+x4+2x2+x1-x2 =∫-1/21/2f(x)(x2+2) dx=π6-14