Download App

DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS1 Mark
MCQ
The solution of the differention equation $\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^{2}+\text{xy}+\text{y}^{2}}{\text{x}^{2}}$ is:
  • A
    $\tan^{-1}\big(\frac{\text{x}}{\text{y}}\big)-\log\text{y}+\text{C}$
  • ✓
    $\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)-\log\text{x}+\text{C}$
  • C
    $\tan^{-1}\big(\frac{\text{x}}{\text{y}}\big)=\log\text{x}+\text{C}$
  • D
    $\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)=\log\text{y}+\text{C}$
✓

Answer

Correct option: B.
$\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)-\log\text{x}+\text{C}$
We have,
$\frac{\text{dy}}{\text{dx}}=\frac{\text{x}^{2}+\text{xy}+\text{y}^{2}}{\text{x}^{2}}\ ...(\text{i})$
This is homogenous differential equation.
Let $\text{y}=\text{ux}$
$\Rightarrow \frac{\text{dy}}{\text{dx}}=\text{u}+\text{x}\frac{\text{du}}{\text{dx}}$
Now, putting equation (i),
$\text{u}+\text{x}\frac{\text{du}}{\text{dx}}=\frac{\text{x}^{2}+\text{x}^{2}\text{u}+\text{x}^{2}\text{u}^{2}}{\text{x}^{2}}$
$\Rightarrow \text{u}+\text{x}\frac{\text{du}}{\text{dx}}=1+\text{u}+\text{u}^{2}$
$\Rightarrow \text{x}\frac{\text{du}}{\text{dx}}=1+\text{u}^{2}$
$\Rightarrow \big(\frac{1}{1+\text{u}^{2}}\big)\text{du}=\frac{1}{\text{x}}\text{dx}$
Intergreting both sides, we get
$\int\big(\frac{1}{1+\text{u}^{2}}\big)\text{du}=\int\frac{1}{\text{x}}\text{dx}$
$\Rightarrow \tan^{-1}\text{u}=\log\text{x}+\text{C}$
$\Rightarrow \tan^{-1}\big(\frac{\text{y}}{2}\big)=\log\text{x}+\text{C}$

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

Let a unit vector $\hat{\mathrm{t}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}$ make angles $\frac{\pi}{2}, \frac{\pi}{3}$ and $\frac{2 \pi}{3}$ with the vectors $\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}}, \frac{1}{\sqrt{2}} \hat{\mathrm{j}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}}$ and $\quad \frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{j}} \quad$ respectively. If $\overrightarrow{\mathrm{v}}=\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{j}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}}$, then $|\hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}|^2$ is equal to
→
Let $R$ be a relation on a set $A$ such that $R = {R^{ - 1}}$, then $R$ is
→
The corner points of the feasible region determined by the following system of linear inequalities$:\ 2\text{x}+\text{y}\le10,\ \text{x}+3\text{y}\le15,\ \text{x},\ \text{y}\ge0$ are $(0, 0), (5, 0), (3, 4)$ and $(0, 5).$ Let $Z = px + qy,$ where $p, q > 0.$ Condition on $p$ and $q$ .so that the maximum of $Z$ occurs at both $(3, 4)$ and $(0, 5)$ is:
→
The value of f(0), so that the function $\text{f(x)}=\frac{2-(256-7\text{x})^{\frac{1}{8}}}{(5\text{x}+32)^\frac{1}{5}-2},\text{ x}\neq0$ is continuous everywhere, is given by:
→
Which one of the following statements is not true:
→
If $f(x) = 2x + {\cot ^{ - 1}}x + \log (\sqrt {1 + {x^2}} - x)$, then $f(x)$
→
Let $\alpha$ be a non-zero real number. Suppose $f: \mathrm{R} \rightarrow$ $\mathrm{R}$ is a differentiable function such that $f(0)=2$ and $\lim _{\mathrm{x} \rightarrow-\infty} \mathrm{f}(\mathrm{x})=1$. If $f^{\prime}(\mathrm{x})=\alpha f(x)+3$, for all $\mathrm{x} \in \mathrm{R}$, then $f\left(-\log _e 2\right)$ is equal to . . . . . . . . . 
→
If $\text{P(B)}=\frac{3}{5},\text{P}(\text{A}|\text{B})=\frac{1}{2}$ and $\text{P}(\text{A}\cup\text{B})=\frac{4}{5},$ then $\text{P}(\overline{\text{A}\cap\text{B}})+\text{P}(\overline{\text{A}}\cap\text{B})=$
→
${d \over {dx}}({x^{{{\log }_e}x}}) = $
→
Choose the correct answer from the given four options.
$\text{X}$ $1$ $2$ $3$ $4$
$\text{P}(\text{X})$ $\frac{1}{10}$ $\frac{1}{5}$ $\frac{3}{10}$ $\frac{2}{5}$
For the following probability distribution $E(X^2)$ is equal to:
→