Let y-yx be the solution of the differential equation x-1y-y-e3xx-12- with y0-13- Then- the point x-43 for the curve y-yx is-

(x+1)dydx y=e^3x (x+1)^2 dydx yx+1=e^3x(x+1) I.F=e^ ∫1/x+1dx=e^ log (x+1)=1x+1 ∴ y (1x+1)=∫e^3x(x+1)x+1dx yx+1=∫ e^3x dx yx+1=e^3x3+C ∵ y(0)=13 13=13+C ∴ C=0 ...

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Byju's Answer

Standard XI

Chemistry

Hydrolysis of Amides

Let y=yx be t...

Question

Let $y = y (x)$ be the solution of the differential equation $(x + 1) y^{'} - y = e^{3 x} (x + 1)^{2}$ , with $y (0) = \frac{1}{3}$ . Then, the point $x = - \frac{4}{3}$ for the curve $y = y (x)$ is:

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Solution

$(x + 1) \frac{d y}{d x} - y = e^{3 x} (x + 1)^{2}$
$\frac{d y}{d x} - \frac{y}{x + 1} = e^{3 x} (x + 1)$
$I.F = e^{- \int \frac{1}{x + 1} d x} = e^{- log (x + 1)} = \frac{1}{x + 1}$
$∴ y (\frac{1}{x + 1}) = \int \frac{e^{3 x} (x + 1)}{x + 1} d x$
$\frac{y}{x + 1} = \int e^{3 x} d x$
$\frac{y}{x + 1} = \frac{e^{3 x}}{3} + C$
$∵ y (0) = \frac{1}{3}$
$\frac{1}{3} = \frac{1}{3} + C$
$∴ C = 0$
So: $y = \frac{e^{3 x}}{3} (x + 1)$
$y^{'} = e^{3 x} (x + 1) + \frac{e^{3 x}}{3} = e^{3 x} (x + \frac{4}{3})$
$y^{''} = 3 e^{3 x} (x + \frac{4}{3}) + e^{3 x} = e^{3 x} (3 x + 5)$
$y^{'} = 0$ at $x = \frac{- 4}{3} & y^{''} = e^{- 4} (1) > 0$ at $x = \frac{- 4}{3}$
$\Rightarrow x = \frac{- 4}{3}$ is point of local minima.

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Let y=y(x) be the solution of the differential equation (x+1)y′−y=e3x(x+1)2, with y(0)=13. Then, the point x=−43 for the curve y=y(x) is:

Let $y = y (x)$ be the solution of the differential equation $(x + 1) y^{'} - y = e^{3 x} (x + 1)^{2}$ , with $y (0) = \frac{1}{3}$ . Then, the point $x = - \frac{4}{3}$ for the curve $y = y (x)$ is: