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Worked Examples · Example 7

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

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The problem is a first-order separable ODE. Separating variables and integrating gives y=x2+log⁡∣x∣+Cy = x^2 + \log|x| + C; using the point (1,1)(1,1) fixes C=0C = 0, so the curve is y=x2+log⁡∣x∣y = x^2 + \log|x|.

The key idea here is separation of variables. The differential equation is given in a form where dydy and dxdx are already on opposite sides of the equals sign — but not quite separated, because xx multiplies dydy. That’s a small fix: just divide both sides by xx (and x≠0x \neq 0 is given, so we’re safe). Once dydy is alone, the right-hand side becomes a function of xx only, and we can integrate both sides directly.

Why does this work? Because the equation is of the form dydx=f(x)\frac{dy}{dx} = f(x) — there’s no yy on the right. That’s the simplest possible case of separation: you don’t even need to rearrange terms involving yy. Just integrate.

Let’s go step by step.

  1. Rewrite the equation in separated form We have x dy=(2x2+1) dxx\,dy = (2x^2 + 1)\,dx. Divide through by xx (allowed since x≠0x \neq 0):

dy=2x2+1x dx.dy = \frac{2x^2 + 1}{x}\,dx.

Simplify the fraction:

2x2+1x=2x+1x.\frac{2x^2 + 1}{x} = 2x + \frac{1}{x}.

So the equation becomes

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

  1. Integrate both sides The left side integrates to yy (plus a constant). The right side is a sum of simple terms:

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

This gives

y=x2+log⁡∣x∣+C,y = x^2 + \log|x| + C,

where CC is the constant of integration.

Note

The absolute value inside the logarithm is important because xx can be negative (though the given point has x=1x=1, the general solution must hold for all x≠0x \neq 0). In many Indian exam contexts, they write log⁡x\log x assuming x>0x>0, but the safer form is log⁡∣x∣\log|x|.

  1. Use the given point to find CC …

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