Methods of solving first order first degree differential equations

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Question

$If for x \geq 0, y = y (x) is the solution of the differential equation,$
$(x + 1) d y = ((x + 1)^{2} + y - 3) d x, y (2) = 0, then y (3) is equal to$

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Solution

$\begin{array}{l} \frac{d y}{d x} = (1 + x) + (\frac{y - 3}{1 + x}) \\ \frac{d y}{d x} - \frac{1}{(1 + x)} y = (1 + x) - \frac{3}{(1 + x)} \\ I.F. = e^{- \int \frac{1}{(1 + x)} d x} = \frac{1}{(1 + x)} \end{array}$
$\begin{array}{l} So, the solution will be \frac{y}{1 + x} = \int [1 - \frac{3}{(1 + x)^{2}}] d x \\ \frac{y}{1 + x} = x + 3 (1 + x)^{- 1} + c \\ y = (1 + x) [x + \frac{3}{(1 + x)} + c] \\ At x = 2, y = 0 \\ 0 = 3 (2 + 1 + c) \Rightarrow c = - 3 \end{array}$
$Hence the solution is y = (1 + x) (x + \frac{3}{1 + x} - 3)$
$S u b s t i t u t e x = 3 I t g i v e s$
$\begin{matrix} y = (1 + 3) (3 + \frac{3}{1 + 3} - 3) \\ = 4 (\frac{3}{4}) \\ = 3 \end{matrix}$

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Methods of solving first order first degree differential equations

Remember concepts with our Masterclasses.

If for x≥0,y=y(x) is the solution of the differential equation, (x+1)dy=(x+1)2+y-3dx,y(2)=0, then y(3) is equal to

dydx=(1+x)+y-31+xdydx-1(1+x)y=(1+x)-3(1+x) I.F. =e-∫1(1+x)dx=1(1+x) So, the solution will be y1+x=∫1-3(1+x)2dxy1+x=x+3(1+x)-1+cy=(1+x)x+3(1+x)+c At x=2,y=00=3(2+1+c)⇒c=-3 Hence the solution is y=(1+x)x+31+x-3Substitute x=3 It givesy=1+33+31+3−3=434=3

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$If for x \geq 0, y = y (x) is the solution of the differential equation,$
$(x + 1) d y = ((x + 1)^{2} + y - 3) d x, y (2) = 0, then y (3) is equal to$