Differential equation having solution y=A x+B^3 is of order - Vidyadip

MCQ

Differential equation having solution $y=A x+B^3$ is of order

A
3
✓
2
C
1
D
not defined

✓

Answer

Correct option: B.

2

(b) : Given solution contanty two arbitrary constant.
$\therefore \quad$ Order differential equation is 2 .

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→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

If $\int_{}^{} {\frac{1}{{(\sin x + 4)(\sin x - 1)}}\;dx = A\frac{1}{{\tan \frac{x}{2} - 1}} + B{{\tan }^{ - 1}}(f(x))} + C$, then

If $\text{A}= \begin{bmatrix} 1 &\text{amp; } 2 &\text{amp;} 3\end{bmatrix},$ then order is:

3 × 1
1 × 3
2 × 3
None of these

A coin is tossed $m + n$ times, where $m \ge n.$ The probability of getting at least $m$ consecutive heads is

$\left|\frac{120}{\pi^3} \int_0^\pi \frac{x^2 \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x\right|$ is equal to....................

A box contains $100$ tickets numbered $1, 2 ...... 100$. Two tickets are chosen at random. It is given that the maximum number on the two chosen tickets is not more than $10$. The minimum number on them is $5$ with probability

Let $y=y(x), x>1$, be the solution of the differential equation $(x-1) \frac{d y}{d x}+2 x y=\frac{1}{x-1}$, with $y(2)=\frac{1+e^{4}}{2 e^{4}}$. If $y(3)=\frac{e^{\alpha}+1}{\beta e^{\alpha}}$. then the value of $\alpha+\beta$ is equal to

If $\ln \left( {\left( {e - 1} \right){e^{xy}} + {x^2}} \right) = {x^2} + {y^2}$ , then ${\left. {\frac{{dy}}{{dx}}} \right|_{\left( {1,0} \right)}}$ is

Let $\lambda \in Z, \vec{a}=\lambda \hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}$. Let $\overrightarrow{ c }$ be a vector such that $(\vec{a}+\vec{b}+\vec{c}) \times \vec{c}=\overrightarrow{0}, \vec{a} \cdot \vec{c}=-17$ and $\vec{b} \cdot \vec{c}=-20$ Then $|\overrightarrow{ c } \times(\lambda \hat{i}+\hat{j}+\hat{ k })|^2$ is equal to

The projection of the join of the two points (1, 4, 5), (6, 7, 2) on the line whose d.ss are (4, 5, 6) is:

If $f(x) = {x^2} + 1$, then ${f^{ - 1}}(17)$ and ${f^{ - 1}}( - 3)$ will be