Download App

Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
Solve the following differential equation:
$(\text{x}+2\text{y})\text{dx}-(2\text{x}-\text{y})\text{dy}=0$
✓

Answer

$(\text{x}+2\text{y})\text{dx}-(2\text{x}-\text{y})\text{dy}=0$$\Rightarrow\ \frac{\text{dy}}{\text{dx}}=\frac{\text{x}+2\text{y}}{2\text{x}-\text{y}}$
This is a homogeneous differential equation. Putting y = vx and $\frac{\text{dy}}{\text{dx}}=\text{v + x}\frac{\text{dv}}{\text{dx}}$, we get $\text{v + x}\frac{\text{dv}}{\text{dx}}=\frac{\text{x}+2\text{vx}}{2\text{x}-\text{vx}}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+2\text{v}}{2-\text{v}}-\text{v}$ $\Rightarrow\ \text{x}\frac{\text{dv}}{\text{dx}}=\frac{1+\text{v}^2}{2-\text{v}}$ $\Rightarrow\ \frac{2-\text{v}}{1+\text{v}^2}\text{dv}=\frac{1}{\text{x}}\text{dx}$ Integrating both sides, we get $\int\frac{2-\text{v}}{1+\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}\ \dots(1)$ $\Rightarrow\ \int\frac{2}{1+\text{v}^2}\text{dv}-\int\frac{\text{v}}{1+\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$ $\Rightarrow\ \int\frac{2}{1+\text{v}^2}\text{dv}-\frac{1}2\int\frac{2\text{v}}{1+\text{v}^2}\text{dv}=\int\frac{1}{\text{x}}\text{dx}$ $\Rightarrow\ 2\tan^{-1}\text{v}-\frac{1}2\log|1+\text{v}^2|=\log|\text{x}|+\log\text{C}$ $\Rightarrow\ 2\tan^{-1}\text{v}=\log|\text{x}|+\log\text{C}+\log\Big|(1+\text{v}^2)^{\frac{1}2}\Big|$ $\Rightarrow\ 2\tan^{-1}\text{v}=\log\Big|\text{Cx}\sqrt{1+\text{v}^2}\Big|$ $\Rightarrow\ \Big|\text{Cx}\sqrt{1+\text{v}^2}\Big|=\text{e}^{2\tan^{-1}\text{v}}$ Putting $\text{v}=\frac{\text{y}}{\text{x}}$, we get $\Rightarrow\ \Bigg|\text{Cx}\sqrt{1+\Big(\frac{\text{y}}{\text{x}}\Big)^2}\Bigg|=\text{e}^{2\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)}$ $\Rightarrow\ \text{C}\sqrt{\text{x}^2+\text{y}^2}=\text{e}^{2\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)}$ Hence, $\sqrt{\text{x}^2+\text{y}^2}=\text{Ke}^{2\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)}$ is the required solution.

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 "Solve the Following Question.(5 Marks)" from Differential Equations→Differential Equations — all question types→Maths STD 12 Science question papers→Maharashtra Board STD 12 Science question papers→Maharashtra Board question paper generator→

Similar questions

Evaluate : $\int \frac{1}{3+2 \sin ^2 x+5 \cos ^2 x} \cdot d x$
→
Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}=\frac{\text{y}(\text{x}+2\text{y})}{\text{x}(2\text{x}+\text{y})},\text{y}(1)=2$
→
If $(\sin\text{x})^{\text{y}}=\text{x}+\text{y},$ prove that $\frac{\text{dy}}{\text{dx}}=\frac{1-(\text{x}+\text{y})\text{y}\cot\text{x}}{(\text{x}+\text{y})\log\sin\text{x}-1}$
→
An urn contains 4 red and 3 blue balls. Find the probability distribution of the number of blue balls in a random draw of 3 balls with replacement.
→
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:
$\text{f}(\text{x})=\sin\text{x}-\cos(\text{x}), 0\leq\text{x}\leq2\pi$
→
If $\text{u}=\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ and $\text{v}=\tan^{-1}\Big(\frac{2\text{x}}{1-\text{x}^2}\Big),$ where -1 < x < 1 then write the value of $\frac{\text{dy}}{\text{dx}}.$
→
Evaluate the following integrals:$\int\limits^{\frac{\pi}{2}}_0\big(2\log\cos\text{x}-\log\sin2\text{x}\big)\text{dx}$
→
Using differentials, find the approximate values of the following:
$\sqrt{49.5}$
→
Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}-3\text{y}\cot\text{x}=\sin2\text{x},\text{ y}=2,\text{ when x}=\frac{\pi}{2}$
→
Show that the following system of linear equations is consistent and also find solutions:
$5x +3y + 7z = 4$
$3x + 26y + 2z = 9$
$7x + 2y + 10z = 5$
→