Download App

STD 12 - 9. differential equations — Mathematics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceMathematicsSTD 12 - 9. differential equations1 Mark
MCQ
A solution of $y = 2x\left( {\frac{{dy}}{{dx}}} \right) + {x^2}{\left( {\frac{{dy}}{{dx}}} \right)^4}$ is
  • A
    $y = 2{C^{\frac{1}{2}}}{x^{\frac{1}{4}}} + C$
  • B
    $y = 2\sqrt C {x^2} + {C^2}$
  • C
    $y = 2\sqrt C \left( {x + 1} \right)$
  • ✓
    $y = 2\sqrt {Cx}  + {C^2}$
✓

Answer

Correct option: D.
$y = 2\sqrt {Cx}  + {C^2}$
d
Writing $\mathrm{P}=\frac{\mathrm{dy}}{\mathrm{dx}}$ and differentiating w.r.t. $\mathrm{x} ;$ we have

$\mathrm{P}=2 \mathrm{P}+2 \mathrm{x} \frac{\mathrm{dP}}{\mathrm{dx}}+2 \mathrm{xP}^{4}+4 \mathrm{P}^{3} \mathrm{x}^{2} \frac{\mathrm{dP}}{\mathrm{dx}}$

$\Rightarrow 0=\mathrm{P}\left(1+2 \mathrm{xP}^{3}\right)+2 \mathrm{x} \frac{\mathrm{dP}}{\mathrm{dx}}\left(1+2 \mathrm{P}^{3} \mathrm{x}\right)$

$\Rightarrow \mathrm{P}+2 \mathrm{x} \frac{\mathrm{dP}}{\mathrm{dx}}=0 \Rightarrow 2 \frac{\mathrm{dP}}{\mathrm{p}}=-\frac{\mathrm{dx}}{\mathrm{x}}$

$\Rightarrow 2 \log \mathrm{P}+\log \mathrm{x}=$ constant

$\Rightarrow \mathrm{P}^{2} \mathrm{x}=\mathrm{C} \Rightarrow \mathrm{P}=\sqrt{\mathrm{C} / \mathrm{x}}$

Substituting this value in the given equation, we

get $y=2 \sqrt{C x}+C^{2}$

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 STD 12 - 9. differential equations→STD 12 - 9. differential equations — all question types→Mathematics STD 12 Science question papers→CBSE Board STD 12 Science question papers→CBSE Board question paper generator→

Similar questions

If $\begin{bmatrix}\text{r}+4&\text{amp; 6}\\3&\text{amp; 3}\end{bmatrix}=\begin{bmatrix}{5}&\text{amp;}\text{ r}+5\\\text{r+2}&\text{amp; 4}\end{bmatrix}$ then $\text{r}=$
→
If $f(x) = \left\{ \begin{gathered} \,[x]\, + \,[ - x],\,\,x \ne 2 \hfill \\ \,\,\,\,\,\,\,\,\,\lambda \,\,\,\,\,\,\,\,\,,\,\,x = \,2\,\,\,\, \hfill \\  \end{gathered}  \right.,$ then $f$ is continuous at $x = 2,$ provided $\lambda$ is (where $[.]$ is $G.I.F.$ )
→
Find the principal value of $\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right)$
→
The number of $3 \times 3$ non- singular matrices, with four entries as $1$ and all other entries as $0$, is 
→
The integral $\int {x\,{{\cos }^{ - 1}}\,\left( {\frac{{1 - {x^2}}}{{1 + {x^2}}}} \right)dx} \,\left( {x > 0} \right)$ is equal to
→
Choose the correct answer from the given four options.The domain of the function $\cos ^{-1}(2 x-1)$ is:
→
For the $\text{LPP};$ maximise $z = x + 4y$ subject to the constraints $\text{x}+2\text{y}\leq2, \text{x}+2\text{y}\geq8,\text{x},\text{y}\geq0.$
→
Choose the correct answer from the given four options.
Corner points of the feasible region for an LPP are (0, 2), (3, 0), (6, 0), (6, 8) and (0, 5).
Let F = 4x + 6y be the objective function.
The Minimum value of F occurs at.
→
The corner point of the feasible region determined by the system of linear constraints are (0, 0), (0, 40), (20, 40), (60, 20), (60, 0). The objective function is Z = 4x + 3y. Compare the quantity in Column A and Column B.
Column A
Column B
Maximum of Z
325
→
Let $\vec{a}=2 \hat{i}+\hat{j}-2 \hat{k}$ and $\vec{b}=\hat{i}+\hat{j} .$ If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c}=|\vec{c}|,|\vec{c}-\vec{a}|=2 \sqrt{2}$ and the angle between $(\vec{a} \times \vec{b})$ and $\vec{c}$ is $\frac{\pi}{6}$, then the value of $|(\vec{a} \times \vec{b}) \times \vec{c}|$ is:
→