MCQ
The solution of the equation $\frac{d y}{d x}+2 x=e^{3 x}$ is :
  • $y+x^2=\frac{1}{3} e^{3 x}+c$
  • B
    $y-x^2=\frac{1}{3} e^{3 x}+c$
  • C
    $y+x^2=e^{3 x}+c$
  • D
    $y-x^2=e^{3 x}+c$

Answer

Correct option: A.
$y+x^2=\frac{1}{3} e^{3 x}+c$
(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

Similar questions

The function f : A → B defined by f(x) = -x2 + 6x- 8 is a bijection if,
  1. $\text{A}=(-\infty,3]$ and $\text{B}=(-\infty,1]$
  2. $\text{A}=[-3,\infty)$ and $\text{B}=(-\infty,1]$
  3. $\text{A}=(-\infty,3]$ and $\text{B}=[1,\infty)$
  4. $\text{A}=[3,\infty)$ and $\text{B}=[1,\infty)$
The vertices $B$ and $C$ of a $\Delta ABC$ lie on the line, $\frac{{x + 2}}{3} = \frac{{y - 1}}{0} = \frac{z}{4}$ such that $BC = 5\, units$. Then the area (in $sq. units$) of this triangle, given that the point $A\, (1, -1, 2)$ is
The angle between the pair of lines with direction ratios $(1, 1, 2)$ and $(\sqrt 3 - 1, - \sqrt 3 - 1,4)$ is ......... $^o$
If $A=\left[\begin{array}{ccc}-2 & 0 & 0 \\ 0 & -2 & 0 \\ 0 & 0 & -2\end{array}\right]$, then the value of $|\operatorname{adj} A|$ is
Number of solution of the equation $ 3tanx + x^3 = 2 $  in $ \left( {0,\frac{\pi }{4}} \right)$ is
If $A$ and $B$ are $2 \times 2$ matrices, then which of the following is true?
The value of $\sin \left( {2{{\tan }^{ - 1}}\left( {\frac{1}{3}} \right)} \right) + \cos ({\tan ^{ - 1}}2\sqrt 2 ) = $
The function $\text{f}:[0,\infty)\rightarrow\ \text{R}$ given by $\text{f(x)}=\frac{\text{x}}{\text{x}+1}$ is:
  1. One-one and onto.
  2. One-one but not onto.
  3. Onto but not one-one.
  4. Onto but not one-one.
$\int {\frac{1}{{{x^2} - 1}}} \,\,\ln \,\frac{{x - 1}}{{x + 1}}dx$ equals :
Let $f(x)$ be a positive function such that the area bounded by $y=f(x), y=0$ from $x=0$ to $x=a>0$ is $\mathrm{e}^{-\mathrm{a}}+4 \mathrm{a}^2+\mathrm{a}-1$. Then the differential equation, whose general solution is $y=c_1 f(x)+c_2$, where $c_1$ and $c_2$ are arbitrary constants, is :