The solution of the equation d y d x +2 x=e^ 3 x is : - Vidyadip

Question

The solution of the equation $\frac{d y}{d x}+2 x=e^{3 x}$ is :

✓

Answer

(A)
$
\begin{aligned}
& \\
\Rightarrow \quad \frac{d y}{d x}+2 x & =e^{3 x} \\
\frac{d y}{d x} & =e^{3 x}-2 x \\
d y & =\left(e^{3 x}-2 x\right) d x \\
\therefore \quad \int d y & =\int\left(e^{3 x}-2 x\right) d x \\
y & =\frac{1}{3} e^{3 x}-\frac{2 x^2}{2}+C \\
y & =\frac{1}{3} e^{3 x}-x^2+C \\
\Rightarrow \quad y+x^2 & =\frac{1}{3} e^{3 x}+C
\end{aligned}
$
Hence the correct choice is (A).

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 "M.C.Q (1 Marks)" 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

The function $f(x)=|x|$ is

The value of $\int\frac{\text{d}(\text{x}^2+1)}{\sqrt{\text{x}^2+2}}$ is:

$2\sqrt{\text{x}^2+2}+\text{c}$
$\sqrt{\text{x}^2+2}+\text{c}$
$\text{x}\sqrt{\text{x}^2+2}+\text{c}$
${4}\sqrt{\text{x}^2+2}+\text{c}$

Let $X = \{-1, 0, 1\}, Y = \{0, 2\}$ and a function $f : X \rightarrow Y$ defiend by $y = 2x^4,$ is$:$

The maximum value of Z = 3x + 4y subjected to constraints $\text{x}+\text{y}\leq4,\text{x}\geq0$ and $\text{y}\geq0$ is:

12
14
16
None of the above

Area of curve explained in the passage from $0 \text{ to}\frac{\pi}{2}\text{ is:}$

$\frac{1}{3}\text{ sq.}\text{ unit}$
$\frac{1}{2}\text{ sq.}\text{ unit}$
$1\text{ sq.}\text{ unit}$
$2\text{ sq.}\text{ units}$

A relation R is defined from {2, 3, 4, 5} to {3, 6, 7, 10} by : xRy ⇔ x is relatively prime to y. Then, domain of R is:

{2, 3, 5}
{3, 5}
{2, 3, 4}
{2, 3, 4, 5}

Area bounded by curve $y^2=4 x, y$-axis and line $y=3$ is:

The distinct linear functions that map $[-1, 1]$ onto $[0, 2]$ are:

If $\text{f(x)}=\text{a}|\sin\text{x}|+\text{be}^{|\text{x}|}+\text{c|x|}^3$and if f(x) is differentiable at x = 0, then:

$\text{a}=\text{b}=\text{c}=0$
$\text{a}=0,\text{b}=0;\text{c}\in\text{R}$
$\text{b}=\text{c}=0,\text{a}\in\text{R}$
$\text{c}=0,\text{a}=0,\text{b}\in\text{R}$

If $\text{f(x)}=\begin{cases}\frac{\sin(\cos\text{x})-\cos\text{x}}{(\pi-2\text{x})^2},&\text{x}\neq\frac{\pi}{2}\\\text{k},&\text{x}=\frac{\pi}{2}\end{cases}$ is continuous at $\text{x}=\frac{\pi}{2},$ then k is equal to:

0
$\frac{1}{2}$
1
-1