Solve the following differential equation: dy dx -y=xe^ x - Vidyadip

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 "Solve the Following Question.(5 Marks)" from Differential Equations→Differential Equations — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

Show that x = 2 is a root of the equation $\begin{vmatrix}\text{x}&-6&-1\\2&-3\text{x}&\text{x}-3\\-3&2\text{x}&\text{x}+2\end{vmatrix}=0$ and solve it completely.

Solve the following systems of linear equations by cramer's rule:

Solve the following determinant equations:
$\begin{vmatrix}\text{x}+\text{a}&\text{x}&\text{x}\\\text{x}&\text{x}+\text{a}&\text{x}\\\text{x}&\text{x}&\text{x}+\text{a}\end{vmatrix}=0,\text{a}\neq0$

Using intergation, find the area of the bounded by the triangle whose vertices are (-1, 2), (1, 5) and (3, 4).

Without expanding, show that the values of the following determinant are zero:
$\begin{vmatrix}1&\text{a}&\text{a}^2-\text{bc}\\1&\text{b}&\text{b}^2-\text{ac}\\1&\text{c}&\text{c}^2-\text{ab} \end{vmatrix}$

If $\text{A}=\begin{bmatrix}1&2&2\\2&1&2\\2&2&1\end{bmatrix},$ then prove that $A^2 - 4A - 5I = 0.$

If $\vec{\text{a}},\vec{\text{b}}$ are two non-collinear vectors, prove that the points with position vectors $\vec{\text{a}}+\vec{\text{b}},\ \vec{\text{a}}-\vec{\text{b}}$ and $\vec{\text{a}}+\lambda\vec{\text{b}}$ are collinear for all real values of $\lambda$.

If $\text{f(x)}=\frac{\tan\big(\frac{\pi}{4}-\text{x}\big)}{\cot2\text{x}}$ for $\text{x}\neq\frac{\pi}{4},$ find the value which can be assigned to f(x) at $\text{x}=\frac{\pi}{4}$ so that the function f(x) becomes continuous every where in $\Big[0,\frac{\pi}{2}\Big]$

If $\begin{vmatrix}\text{p}&\text{q}&\text{c}\\\text{a}&\text{q}&\text{c}\\\text{a}&\text{b}&\text{r} \end{vmatrix}=0,$ find the value of $\frac{\text{p}}{\text{p}-\text{a}}+\frac{\text{q}}{\text{q}-\text{b}}+\frac{\text{r}}{\text{r}-\text{c}},$ $\text{p}\neq\text{a},\text{q}\neq\text{b},\text{r}\neq\text{c}.$

Evaluate the following integrals:$\int\limits^2_{-2}\frac{3\text{x}^3+2|\text{x}|+1}{\text{x}^2+|\text{x}|+1}\text{ dx}$