Question

If for $\mathrm{x}\ge 0$,$\mathrm{y}=\mathrm{y}\left(\mathrm{x}\right)$ is the solution of the differential equation $(1+\mathrm{x})\mathrm{dy}=\left[{\left(1+\mathrm{x}\right)}^2+\mathrm{y}-3\right]\mathrm{dx},\mathrm{y}\left(2\right)=0,$then $\mathrm{y}\left(3\right)$ is equal to:

Answer 1

Explanation for the correct option

Step 1: Finding the value of $\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}$

$$\begin{gathered} \therefore (1+\mathrm{x})\mathrm{dy}=\left[{\left(1+\mathrm{x}\right)}^2+\mathrm{y}-3\right]\mathrm{dx} \\ \Rightarrow (1+\mathrm{x})\frac{\mathrm{dy}}{\mathrm{dx}}=\left[{\left(1+\mathrm{x}\right)}^2+\mathrm{y}-3\right] \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{(1+\mathrm{x})}^2}{(1+\mathrm{x})}+\frac{(\mathrm{y}-3)}{(1+\mathrm{x})} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\left(1+\mathrm{x}\right)+\frac{(\mathrm{y}-3)}{(1+\mathrm{x})} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}-\frac{\mathrm{y}}{1+\mathrm{x}}=(1+\mathrm{x})-\frac{3}{(1+\mathrm{x})} \end{gathered}$$

Step 2: Finding Integration Factor

Integration Factor:

$\begin{array}{rcl}\therefore {\mathrm{e}}^{-\int \frac{1}{1+\mathrm{x}}\mathrm{dx}}& =& {\mathrm{e}}^{-\log (1+\mathrm{x})}\\ & =& \frac{1}{1+\mathrm{x}}\end{array}$

$$\begin{gathered} \therefore \mathrm{y}\times \mathrm{I}.\mathrm{F}=\int \mathrm{I}.\mathrm{F}\left(\mathrm{Q}\right) \\ \Rightarrow \mathrm{y}\times \frac{1}{1+\mathrm{x}}=\int \frac{1}{1+\mathrm{x}}\left((1+\mathrm{x})-\frac{3}{(1+\mathrm{x})}\right)\mathrm{dx} \\ \Rightarrow \frac{\mathrm{y}}{1+\mathrm{x}}=\int 1-\frac{3}{{(1+\mathrm{x})}^2}\mathrm{dx} \\ \Rightarrow \frac{\mathrm{y}}{1+\mathrm{x}}=\mathrm{x}-(-1)\frac{3}{1+\mathrm{x}}+\mathrm{C} \\ \Rightarrow \mathrm{y}=(1+\mathrm{x})\left(\mathrm{x}+\frac{3}{1+\mathrm{x}}+\mathrm{C}\right) \end{gathered}$$

Step 3: Finding $C$ by equating $y\left(2\right)=0$

$\therefore \mathrm{y}\left(2\right)=0$

$$\begin{gathered} \therefore (1+2)\left(2+\frac{3}{1+2}+\mathrm{C}\right)=0 \\ \Rightarrow 3(2+1+\mathrm{C})=0 \\ \Rightarrow 6+3+3\mathrm{C}=0 \\ \Rightarrow 3\mathrm{C}=-9 \\ \Rightarrow \mathrm{C}=-3 \end{gathered}$$

Step 3: Finding $\mathit{y}\mathbf{\left(}\mathbf{3}\mathbf{\right)}$

$\begin{array}{rcl}\therefore y\left(3\right)& =& 0\end{array}$

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Hence, the value of $\mathbf{y}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{3}$