First slide

Methods of solving first order first degree differential equations

Question

$\begin{array}{l} y (x) is a function which satisfies the differential equation \frac{d y}{d x} = (y + 1) [(y + 1) e^{\frac{x^{2}}{2}} - x] \\ such that y (2) = 0, then the value of y^{'} (1) = \end{array}$

Moderate

A

$- \frac{2 e^{2}}{{(1 + e)}^{2}}$
B

$\frac{- e^{\frac{3}{2}}}{{(e^{2} + 1)}^{2}}$
C

$\frac{e^{\frac{5}{2}}}{{(1 + e^{2})}^{2}}$
D

$\frac{5 e^{\frac{1}{2}}}{{(e^{2} + 1)}^{2}}$

Solution

$\begin{array}{l} The given differential equation is \frac{d y}{d x} = - x (y + 1) + {(y + 1)}^{2} e^{\frac{x^{2}}{2}} \\ This can be written as \\ \frac{d y}{d x} + x (y + 1) = {(y + 1)}^{2} e^{\frac{x^{2}}{2}} \\ \frac{1}{y + 1} \frac{d y}{d x} + x = (y + 1) e^{\frac{x^{2}}{2}} \\ substitute \frac{1}{y + 1} = t \Rightarrow - \frac{1}{{(y + 1)}^{2}} \frac{d y}{d x} = \frac{d t}{d x} \\ Hence the given differential equation is \frac{d t}{d x} - t x = e^{\frac{x^{2}}{2}} \\ This is a linear differential equation \\ Integrating factor is e^{- \frac{x^{2}}{2}} \\ Solution is t e^{- \frac{x^{2}}{2}} = \int 1 d x \\ \frac{1}{y + 1} e^{- \frac{x^{2}}{2}} = - x + c \end{array}$
$\begin{array}{l} Given y (2) = 0 \\ hence, e^{- 2} = - 2 + c \\ c = \frac{1}{e^{2}} + 2 \\ Solution is \\ \frac{1}{y + 1} e^{- \frac{x^{2}}{2}} = - x + \frac{1}{e^{2}} + 2 \\ (- x + 2 + \frac{1}{e^{2}}) (y + 1) e^{\frac{x^{2}}{2}} = 1 \\ plug in x = 1 \\ (1 + \frac{1}{e^{2}}) e^{\frac{1}{2}} = \frac{1}{y + 1} \\ y + 1 = \frac{e^{- \frac{1}{2}}}{1 + e^{- 2}} \end{array}$
$S u b s t i t u t e (y + 1) i n t h e g i v e n d i f f e r e n t i a l e q u a i t o n$
${[\frac{d y}{d x}]}_{x = 1} = \frac{e^{- \frac{1}{2}}}{1 + e^{- 2}} (\frac{1 - 1 - e^{- 2}}{1 + e^{- 2}}) = \frac{- e^{- \frac{5}{2}}}{{(1 + e^{- 2})}^{2}} = \frac{- e^{\frac{3}{2}}}{{(e^{2} + 1)}^{2}}$

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