Model Paper 1 — Applied Maths STD 12 Science — Question
CBSE BoardEnglish MediumSTD 12 ScienceApplied MathsModel Paper 13 Marks
Question
Solve the initial value problem: $e^{\frac{d y}{d x}}=x+1 ; y(0)=3$
✓
Answer
The given differential equation is, $ e^{\frac{d y}{d x}}=x+1 $ Taking log on both sides, we get, $ \begin{array}{l} \frac{d y}{d x} \log e=\log (x+1) \\ \Rightarrow \frac{d y}{d x}=\log (x+1) \\ \Rightarrow dy=\{\log (x+1)\} dx \end{array} $ Integrating both sides, we get $\int d y=\int\{\log (x+1) d x$ $\Rightarrow$ y = $\int \frac{1}{I I}$ $\times$ log$(\underset{I}{x}+1)$dx $\begin{array}{l}\Rightarrow y =\log ( x +1) \int 1 dx -\int\left[\frac{d}{d x}(\log x +1) \int 1 dx \right] dx \\ \Rightarrow y = x \log ( x +1)-\int \frac{x}{x+1} d x \\ \Rightarrow y = x \log ( x +1)-\int\left(1-\frac{1}{x+1}\right) d x \\ \Rightarrow y = x \log ( x +1)- x +\log ( x +1)+ C \ldots( i )\end{array}$ It is given that y(0) = 3 $\begin{array}{l}\therefore 3=0 \times \log (0+1)-0+\log (0+1)+C \\ \Rightarrow C=3\end{array}$ Substituting the value of C in (i), we get $ \begin{array}{l} y=x \log (x+1)+\log (x+1)-x+3 \\ \Rightarrow y=(x+1) \log (x+1)-x+3 \end{array} $ Hence, $y=(x+1) \log (x+1)-x+3$ is the solution to the given differential equation.
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