The solution curve of the differential equation,
(1 + e-x)(1 + y2)${{dy} \over {dx}}$ = y2,
which passes through
the point (0, 1), is :
Solution
Given (1 + e<sup>-x</sup>)(1 + y<sup>2</sup>)${{dy} \over {dx}}$ = y<sup>2</sup>
<br><br>$\Rightarrow$ $\left( {{{{y^2} + 1} \over {{y^2}}}} \right)dy = {{{e^x}dx} \over {{e^x} + 1}}$
<br><br>Integrating both sides,
<br><br>$$\int {\left( {{{{y^2} + 1} \over {{y^2}}}} \right)dy = } \int {{{{e^x}dx} \over {{e^x} + 1}}} $$<br><br>$\Rightarrow$ $y - {1 \over y} = ln\left| {{e^x} + 1} \right| + c$<br><br>It passes through (0, 1)<br><br>$\Rightarrow$ c = $- ln2$<br><br>$\Rightarrow$ ${y^2} - 1 = yln\left( {{{{e^x} + 1} \over 2}} \right)$<br><br>$\Rightarrow$ ${y^2} = 1 + yln\left( {{{{e^x} + 1} \over 2}} \right)$
About this question
Subject: Mathematics · Chapter: Differential Equations · Topic: Order and Degree
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