Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS3 Marks
Question
Differential equation $\frac{\text{dy}}{\text{dx}}+\text{y}=0,\text{y}(0)=3$ Function $\text{y}=\text{e}^\text{-x}+2$
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Answer
Here, $\text{y}=\text{e}^{\text{x}}+1 ....(1)$
Differentiating it with respect to $x,$
$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}}$
$\frac{\text{dy}}{\text{dx}}=\text{y}-1 ...(2)$
Again, differentiating it with respect to $x,$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}=\frac{\text{dy}}{\text{dx}}$
$\frac{\text{d}^{2}\text{y}}{\text{dx}^{2}}-\frac{\text{dy}}{\text{dx}}=0$
It is given differential equation.
so,$y = e^{x }+ 1$ is a solution of the equation
put $x - 0$ in equation $(1),$
$\Rightarrow y = e^{0 }+ 1 = 2$
$y(0) = 2$
put $x = 0$ in equation $(2),$
$y' = e^0 = 1$
$y(0) = 1$
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