Download App

Differential Equations — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations4 Marks
Question
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}-\text{y}=\text{xe}^{\text{x}}$
✓

Answer

We have,

$\frac{\text{dy}}{\text{dx}}-\text{y}=\text{xe}^{\text{x}}\ \dots(1)$

Clearly, it is a linear differential equation of the form

$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$

where

$\text{P}=-1$

$\text{Q}=\text{e}^{\text{x}}$

$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$

$=\text{e}^{-\int\text{dx}}$

$=\text{e}^{-\text{x}}$

Multiplying both sides of (1) by e-x, we get

$\text{e}^{-\text{x}}\Big(\frac{\text{dy}}{\text{dx}}-\text{y}\Big)=\text{xe}^{\text{x}}\text{e}^{-\text{x}}$

$\Rightarrow\ \text{e}^{-\text{x}}\frac{\text{dy}}{\text{dx}}-\text{e}^{-\text{x}}\text{y}=\text{x}$

Integrating both sides with respect to x, we get

$\text{e}^{-\text{x}}\text{y}=\int\text{xdx + C}$

$\Rightarrow\ \text{e}^{-\text{x}}\text{y}=\frac{\text{x}^2}{2}+\text{C}$

$\Rightarrow\ \text{y}=\Big(\frac{\text{x}^2}{2}+\text{C}\Big)\text{e}^{\text{x}}$

Hence, $\text{y}=\Big(\frac{\text{x}^2}{2}+\text{C}\Big)\text{e}^{\text{x}}$ 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 "4 Marks" from Differential Equations→Differential Equations — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

By Using properties of definite integrals, evaluate the following integral in Exercise:
$\int^{\pi}_{0}\log(1+\cos\text{x})\text{dx}$
→
Find a unit vector perpendicular to each of the vectors $\overrightarrow{\text{a}}+ \overrightarrow{\text{b}}$ and $\overrightarrow{\text{a}}- \overrightarrow{\text{b}},$ where $\overrightarrow{\text{a}}=3\hat{\text{i}}+2\hat{\text{j}}+2\hat{\text{k}}$ and $\overrightarrow{\text{b}}=\hat{\text{i}}+2\hat{\text{j}}-2\hat{\text{k}}.$
→
Evaluate:

$\int\sqrt{\tan\theta}\text{ d}\theta. $

→
A curve is such that the length of the perpendicular from the origin on the tangent at any point P of the curve is equal to the abscissa of P. Prove that the differential equation of the curve is $\text{y}^{2}-2\text{xy}\frac{\text{dy}}{\text{dx}}-\text{x}^{2}=0$ and hence find the curve.
→
Show that the lines $\frac{5 - x }{-4} =\frac{\text{y}- 7}{4} = \frac{\text{z} + 3}{-5}\text{ and } \frac{{x} - 8}{7} =\frac{2\text{y} - 8}{2} =\frac{\text{z} - 5 }{3}$ are coplanar.
→
If $\text{xy}\log(\text{x}+\text{y})=1,$ prove that $\frac{\text{dx}}{\text{dx}}=-\frac{\text{y}(\text{x}^2\text{y}+\text{x}+\text{y})}{\text{x}(\text{xy}^2+\text{x}+\text{y})}$
→
Find the shortest distance between the following pairs of lines whose vector equation are:
$\vec{\text{r}}=(8+3\lambda)\hat{\text{i}}-(9+16\lambda)\hat{\text{j}}+(10+7\lambda)\hat{\text{k}}$ and $\vec{\text{r}}=15\hat{\text{i}}+29\hat{\text{j}}+5\hat{\text{k}}+\mu\big(3\hat{\text{i}}+8\hat{\text{j}}-5\hat{\text{k}}\big)$
→
Find the equations of the tangent and the normal to the following curves at the indicated points.
$\text{y}=\text{x}^2+4\text{x}+1\text{ at }\text{x}=3$
→
Find the equation of the plane determined by the points A (3, –1, 2), B (5, 2, 4) and C (–1, –1, 6). Also find the distance of the point P (6, 5, 9) from the plane.
→
If $A=\left[\begin{array}{cc}0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0\end{array}\right]$ and I is the identity matrix of order 2 , show that
$I+A=(I-A)\left[\begin{array}{cc}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{array}\right]$
→