Download App

DIFFERENTIAL EQUATIONS — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDIFFERENTIAL EQUATIONS5 Marks
Question
Solve the following initial value problems:
$\text{x}(\text{x}^2+3\text{y}^2)\text{dx}+\text{y}(\text{y}^2+3\text{x}^2)\text{dy}=0,\text{y}(1)=1$
✓

Answer

$\text{x}(\text{x}^2+3\text{y}^2)\text{dx}+\text{y}(\text{y}^2+3\text{x}^2)\text{dy}=0,\text{y}(1)=1$
$\frac{\text{dy}}{\text{dx}}=-\frac{\text{x}(\text{x}^2+3\text{y}^2)}{\text{y}(\text{y}^2+3\text{x}^2)}$
It is a homogeneous equation
put y = vx
$\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$
So,
$\text{v + x}\frac{\text{dv}}{\text{dx}}=-\frac{\text{x}(\text{x}^2+3\text{v}^2\text{x}^2)}{\text{vx}(\text{v}^2\text{x}^2+3\text{x}^2)}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{(1+3\text{v}^2)}{\text{v}(\text{v}^2+3)}-\text{v}$
$\text{x}\frac{\text{dv}}{\text{dx}}=\frac{-1-3\text{v}^2-\text{v}^4-3\text{v}^2}{\text{v}(\text{v}^2+3)}$
$=\frac{-\text{v}^4-6\text{v}^2-1}{\text{v}(\text{v}^2+3)}$
$\frac{\text{v}(\text{v}^2+3)}{\text{v}^4+6\text{v}^2+1}\text{dv}=-\frac{\text{dx}}{\text{x}}$
$\int\frac{4\text{v}^3+12\text{v}}{\text{v}^4+6\text{v}^2+1}\text{dv}=-4\int\frac{\text{dx}}{\text{x}}$
$\log|\text{v}^4+6\text{v}^2+1|=\log\Big|\frac{\text{C}}{\text{x}^4}\Big|$
$|\text{v}^4+6\text{v}^2+1|=\Big|\frac{\text{C}}{\text{x}^4}\Big|\ \dots(\text{i})$
Put y = 1, x = 1
(1+6+1) = C
⇒ C = 8
Put C = 8 in equation (i),
$(\text{y}^4+\text{x}^4+6\text{x}^2\text{y}^2)=8$

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 "5 Marks Questions" 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

Integrate the function in Exercise:
$\frac{1}{\text{x}^{2}(\text{x}^{2}+1)^{\frac{3}{4}}}$
→
The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was $20, 000$ in $1999$ and $25000$ in the year $2004$, what will be the population of the village in $2009$?
→
Find $\frac{d y}{d x}$
$(a) \ \sqrt{x^2+y^2}=\log _e\left(x^2-y^2\right)$
$(b) \ y=x^{x^{x^x} \ldots......\infty}$
$(c) \ x y \log (x+y)=1$
→
Solve the following differential equation
$\text{x}\frac{\text{dy}}{\text{dx}}+1=0;\text{y}(-1)=0$
→
Find the points o local maxima or local minima, if any, of the following functions, using the first derivatives test. Also, find the local maximum or local minimum values, as the case may be:
$\text{f}(\text{x})=\sin2\text{x}-{\text{x}},-\frac{\pi}{2}\leq\text{x}\leq\frac{\pi}{2}$
→
If $\text{y}=\log\sqrt{\text{x}+1}+\sqrt{\text{x}-1},$ show that $\sqrt{\text{x}^2-1}\frac{\text{dy}}{\text{dx}}=\frac{1}{2}\text{y}.$
→
Solve the following systems of linear equations by cramer's rule:
2x - 3y - 4z = 29,
-2x + 5y - z = -15,
3x - y + 5z = -11
→
Solve the following differential equation:
$\Big(1+\text{e}^{\frac{\text{x}}{\text{y}}}\Big)\text{dx}+\text{e}^{\frac{\text{x}}{\text{y}}}\Big(1-\frac{\text{x}}{\text{y}}\Big)\text{dy}=0$
→
Find the angle of intersection of the curves $y^2 = 4ax$ and $x^2 = 4by.$
→
Differentiate the following functions with respect to x:
$\frac{\text{e}^\text{x}\log\text{x}}{\text{x}^2}$
→