Solve the following differential equation: 4 dy dx +8y=5e^ -3x - Vidyadip

Question

Solve the following differential equation:

$4\frac{\text{dy}}{\text{dx}}+8\text{y}=5\text{e}^{-3\text{x}}$

✓

Answer

$4\frac{\text{dy}}{\text{dx}}+8\text{y}=5\text{e}^{-3\text{x}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}+2\text{y}=\frac{5}4\text{e}^{-3\text{x}}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
$\text{P}=2$
$\text{Q}=\frac{5}4\text{e}^{-3\text{x}}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int2\text{dx}}$
$=\text{e}^{2\text{x}}$
Multiplying both sides of (1) by $\text{e}^{2\text{x}},$ we get
$\text{e}^{2\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+2\text{y}\Big)=\frac{5}4\text{e}^{2\text{x}}\text{e}^{-3\text{x}}$
$\Rightarrow\ \text{e}^{2\text{x}}\frac{\text{dy}}{\text{dx}}+2\text{e}^{2\text{x}}\text{y}=\frac{5}4\text{e}^{-\text{x}}$
Integrating both sides with respect to x, we get
$\text{y}\text{e}^{2\text{x}}=\frac{5}4\int\text{e}^{-\text{x}}\text{dx + C}$
$\Rightarrow\ \text{y}\text{e}^{2\text{x}}=-\frac{5}4\text{e}^{-\text{x}}+\text{C}$
$\Rightarrow\ \text{y}=\frac{5}4\text{e}^{-3\text{x}}+\text{C}\text{e}^{-2\text{x}}$
Hence, $\text{y}=\frac{5}4\text{e}^{-3\text{x}}+\text{C}\text{e}^{-2\text{x}}$ 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

Let, X denote the number of colleges where you will apply after your results and P (X = x) denotes your probability of getting admission in x number of colleges. It is given that
$ \text{P(x = }x) = \begin{cases} \text{K}x\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }, & \text{if } x = 0 \text{ or 1}\\ 2\text{k}x\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }, & \text{if } x = 2\\ \text{k} (5 - x)\text{ }\text{ }\text{ }\text{ }, &\text{if } x = 3 \text{ or 4'}\\ 0\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }, & \text{if } x > 4 \end{cases}$
where k is a positive constant. Find the value of k. Also find the probability that you will get admission in (i) exactly one college (ii) at most 2 colleges (iii) at least 2 colleges.

Find the equation of the plane through the points (2, 2, -1) and (3, 4, 2) and parallel to the line whose direction ratios are 7, 0, 6.

Solve the following differential equations:
$\text{x}\frac{\text{dy}}{\text{dx}}+\cot\text{y}=0,$ given that $\text{y}=\frac{\pi}{4},$ when $\text{x}=\sqrt{2}.$

Find the shortest distance between the lines given by $\vec{\text{r}}=(8+3\lambda)\hat{\text{i}}-(9-16\lambda)\hat{\text{j}}+(10+7\lambda)\hat{\text{k}}$ and $\vec{\text{r}}=15\hat{\text{i}}+29\hat{\text{j}}+5\hat{\text{k}}+\mu(3\hat{\text{i}}+8\hat{\text{j}}-5\hat{\text{k}}).$

Find all vectors of magnitude $10\sqrt{3}$ that are perpendicular to the plane of $\hat{\text{i}}+2\hat{\text{j}}+\hat{\text{k}}$ and $-\hat{\text{i}}+3\hat{\text{i}}+4\hat{\text{k}}.$

Find the shortest distance between the lines
$\frac{\text{x}+1}{7}=\frac{\text{y}+1}{-6}=\frac{\text{z}+1}{1}$ and $\frac{\text{x}-3}{1}=\frac{\text{y}-5}{-2}=\frac{\text{z}-7}{1}$

Let $\text{A}=\begin{bmatrix}-1&1&-1\\3&-3&3\\5&5&5\end{bmatrix}$ and $\text{B}=\begin{bmatrix}0&4&3\\1&-3&-3\\-1&4&4\end{bmatrix},$ compute A² - B².

Find the foot of the perpendicular drawn from the point $\hat{\text{i}}+6\hat{\text{j}}+3\hat{\text{k}}$ to the line $\vec{\text{r}}=\hat{\text{j}}+2\hat{\text{k}}+\lambda\big(\hat{\text{i}}+2\hat{\text{j}}+3\hat{\text{k}}\big).$ Also, find the length of the perpendicylar

Using mean value theorem, prove that there is a point on the curve y = 2x² - 5x + 3 between the points A(1, 0) and B(2, 1), where tangent is parallel to the chord AB. Also, find that point.

Express the following matrix as the sum of a symmetric and skew-symmetric matrix and verify your result:
$\text{A}=\begin{bmatrix}3 & -2 &-4\\3 & -2&-5\\-1&-1& 2\end{bmatrix}$