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# Find the Equation of the Curve Passing Through the Point (1, 1) Whose Differential Equation is X Dy = (2x2 + 1) Dx, X ≠ 0. - Mathematics

Sum

Find the equation of the curve passing through the point (1, 1) whose differential equation is x dy = (2x2 + 1) dx, x ≠ 0.

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#### Solution

We have,

$x dy = \left( 2 x^2 + 1 \right)dx$

$\Rightarrow dy = \left( \frac{2 x^2 + 1}{x} \right)dx$

$\Rightarrow dy = \left( 2x + \frac{1}{x} \right)dx$

Integrating both sides, we get

$\int dy = \int\left( 2x + \frac{1}{x} \right)dx$

$\Rightarrow y = x^2 + \log \left| x \right| + C . . . . . . . . . . \left( 1 \right)$

Now the given curve passes through (1, 1)

Therefore, when x = 1, y = 1\]

$\therefore 1 = 1 + 0 + C$

$\Rightarrow C = 0$

Putting the value of C in (1), we get

$y = x^2 + \log\left| x \right|$

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#### APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 68 | Page 147
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