Download App

Differential Equations — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDifferential Equations5 Marks
Question
Solve the following initial value problems:
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=(\text{x}+1)\text{e}^{\text{x}},\text{ y}(1)=0$
✓

Answer

We have,
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=(\text{x}+1)\text{e}^{\text{x}}$
$\Rightarrow\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}}\text{y}=\Big(\frac{\text{x}+1}{\text{x}}\Big)\text{e}^{\text{x}}\ ....(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
Where $\text{P}=-\frac{1}{\text{x}}$ and $\text{Q}=\frac{\text{x}+1}{\text{x}}\text{e}^{-\text{x}}$
$\therefore\text{ I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{-\int\frac{1}{\text{x}}\text{dx}}$
$=\text{e}^{\log\text{x}}$
$=\frac{1}{\text{x}}$
Multiplying both sides of (1) by $\text{I.F.}=\frac{1}{\text{x}},$ we get
$\frac{1}{\text{x}}\Big(\frac{\text{dy}}{\text{dx}}-\frac{1}{\text{x}}\text{y}\Big)=\Big(\frac{\text{x}+1}{\text{x}^2}\Big)\text{e}^{-\text{x}}$
Integrating both sides with respect to x, we get
$\frac{1}{\text{x}}\text{y}=\int\Big(\frac{1}{\text{x}}+\frac{1}{\text{x}^2}\Big)\text{e}^{-\text{x}}\text{dx}+\text{C}$
Putting $\frac{1}{\text{x}}\text{e}^{-\text{x}}=\text{t}$
$\Rightarrow\Big(-\frac{1}{\text{x}}\text{e}^{-\text{x}}-\frac{1}{\text{x}^2}\text{e}^{-\text{x}}\Big)\text{dx}=\text{dt}$
$\Rightarrow\Big(\frac{1}{\text{x}}+\frac{1}{\text{x}^2}\Big)\text{e}^{-\text{x}}\text{dx}=-\text{dt}$
$\therefore\ \frac{1}{\text{x}}\text{y}=\int-\text{dt}+\text{C}$
$\Rightarrow\frac{\text{y}}{\text{x}}=-\text{t}+\text{C}$
$\Rightarrow\frac{\text{y}}{\text{x}}=-\frac{\text{e}^{\text{x}}}{\text{x}}+\text{C}$
$\Rightarrow\text{y}=-\text{e}^{-\text{x}}+\text{Cx}\ ...(2)$
Now,
$\text{y}(1)=0$
$\therefore0=-\text{e}^{-1}+\text{C}$
$\Rightarrow\text{y}=\text{xe}^{-1}-\text{e}^{-\text{x}}$
Hence, $\text{y}=\text{xe}^{-1}-\text{e}^{-\text{x}}$ is the required solution.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from Differential Equations→Differential Equations — all question types→MATHS STD 12 Science question papers→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

Differentiate the following functions with respect to x:
$\frac{\text{e}^\text{x}\sin\text{x}}{(\text{x}^2+2)^3}$
→
$ \text{Find}\ \frac{1}{2}(\text{A}+\text{A}')\ \text{and}\frac{1}{2}(\text{A}-\text{A}'),\ \text{when}\ \text{A}=\begin{bmatrix}0&\text{a}&\text{b}\\-\text{a}&0&\text{c}\\-\text{b}&-\text{c}&0\end{bmatrix}$
→
Find the slopes of the tangent and the normal to the following curves at the indicated points:
$\text{x}^2+3\text{y}+\text{y}^2=5\ \text{at}\ (1,1)$
→
Solve the following LPP graphically:
Manimize Z = 6x + 3y
Subject to the constraints:
$4\text{x}+\text{y}\geq80$
$\text{x}+5\text{y}\geq115$
$3\text{x}+2\text{y}\leq150$
$\text{x}\geq0,\text{y}\geq0$
→
Evaluate $\int\limits_{0}^{2}\text{(x}^{2}+\text{x}+2)\text{dx}$ as limit of sums.
→
Using differentials, find the approximate values of the following:
$\sqrt{37}$
→
Show that the lines $\frac{\text{x}-1}{3}=\frac{\text{y}+1}{2}=\frac{\text{z}-1}{5}$ and $\frac{\text{x}+2}{4}=\frac{\text{y}-1}{3}=\frac{\text{z}+1}{-2}$ do not intersect.
→
Find the area of the bounded by $\text{y}=\sqrt{\text{x}}$ and $y^2 = x.$
→
Maximize $Z = x + y$
Subject to $-2\text{x}+\text{y}\leq1$
$\text{x}\leq2$
$\text{x}+\text{y}\leq3$
$\text{x},\text{y}\geq0$
→
$\text{Find:}\int \frac{\text{x}\ \text{dx}}{(2 + \text{x}^{\text{2}}) (4 + \text{e}^{4\text{}})}$
→