Let us consider a curve, y = f(x) passing through the point ($-$2, 2) and the slope of the tangent to the curve at any point (x, f(x)) is given by f(x) + xf'(x) = x2. Then :
Solution
$y + {{xdy} \over {dx}} = {x^2}$ (given)<br><br>$\Rightarrow {{dy} \over {dx}} + {y \over x} = x$<br><br>If ${e^{\int {{1 \over x}dx} }} = x$<br><br>Solution of DE<br><br>$\Rightarrow y\,.\,x = \int {x\,.\,x\,dx}$<br><br>$\Rightarrow xy = {{{x^3}} \over 3} + {c \over 3}$<br><br>Passes through ($-$2, 2), so<br><br>$-$12 = $-$ 8 + c $\Rightarrow$ c = $-$ 4<br><br>$\therefore$ 3xy = x<sup>3</sup> $-$ 4<br><br>i.e. 3x . f(x) = x<sup>3</sup> $-$ 4
About this question
Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree
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