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Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations2 Marks
Question
Verify that the function $y = \sqrt {1 + {x^2}}$ (explicit or implicit) is a solution of differential equation $y' = \frac{{xy}}{{1 + {x^2}}}$
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Answer

Given: $y = \sqrt {1 + {x^2}} $ …(i)
To prove:y is a solution of the differential equation $y' = \frac{{xy}}{{1 + {x^2}}}$ …(ii)
Proof: From eq. (i),
$y' = \frac{d}{{dx}}\sqrt {1 + {x^2}} = \frac{d}{{dx}}{\left( {1 + {x^2}} \right)^{1/2}}$
$= \frac{1}{2}{\left( {1 + {x^2}} \right)^{ - 1/2}}\frac{d}{{dx}}\left( {1 + {x^2}} \right) = \frac{1}{2}{\left( {1 + {x^2}} \right)^{ - 1/2}}.2x$
$ = \frac{x}{{\sqrt {1 + {x^2}} }}$ …(iii)
Now R.H.S. of eq. (ii)
$ = \frac{{xy}}{{1 + {x^2}}}$
$= \frac{x}{{1 + {x^2}}}\sqrt {1 + {x^2}} $ [From eq. (i)]
$ = \frac{x}{{\sqrt {1 + {x^2}} }} = y'$
$\therefore$ L.H.S. = R.H.S
Hence, y given by eq. (i) is a solution of $y' = \frac{{xy}}{{1 + {x^2}}}$.

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