Let the solution curve y = y(x) of the differential equation
$(4 + {x^2})dy - 2x({x^2} + 3y + 4)dx = 0$ pass through the origin. Then y(2) is equal to _____________.
Answer (integer)
12
Solution
<p>$(4 + {x^2})dy - 2x({x^2} + 3y + 4)dx = 0$</p>
<p>$\Rightarrow {{dy} \over {dx}} = \left( {{{6x} \over {{x^2} + 4}}} \right)y + 2x$</p>
<p>$\Rightarrow {{dy} \over {dx}} - \left( {{{6x} \over {{x^2} + 4}}} \right)y = 2x$</p>
<p>$I.F. = {e^{ - 3\ln ({x^2} + 4)}} = {1 \over {{{({x^2} + 4)}^3}}}$</p>
<p>So ${y \over {{{({x^2} + 4)}^3}}} = \int {{{2x} \over {{{({x^2} + 4)}^3}}}dx + c}$</p>
<p>$\Rightarrow y = - {1 \over 2}({x^2} + 4) + c{({x^2} + 4)^3}$</p>
<p>When x = 0, y = 0 gives $c = {1 \over {32}}$,</p>
<p>So, for x = 2, y = 12</p>
About this question
Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree
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