Download App

Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
Solve the following differential equation:
$(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}+\text{y}=\text{e}^{\tan^{-1}\text{x}}$
✓

Answer

We have,
$(1+\text{x}^2)\frac{\text{dy}}{\text{dx}}+\text{y}=\text{e}^{\tan^{-1}\text{x}}$
$\Rightarrow\ \frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{1+\text{x}^2}=\frac{\text{e}^{\tan^{-1}\text{x}}}{1+\text{x}^2}\ \dots(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
where
$\text{P}=\frac{1}{1+\text{x}^2}$
$\text{Q}=\frac{\text{e}^{\tan^{-1}\text{x}}}{1+\text{x}^2}$
$\therefore$ I.F. $=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\frac{1}{1+\text{x}^2}\text{dx}}$
Multiplying both sides of (1) by $\text{e}^{\tan^{-1}\text{x}},$ we get
$\text{e}^{\tan^{-1}\text{x}}\Big(\frac{\text{dy}}{\text{dx}}+\frac{\text{y}}{1+\text{x}^2}\Big)=\text{e}^{\tan^{-1}\text{x}}\frac{\text{e}^{\tan^{-1}\text{x}}}{1+\text{x}^2}$
$\Rightarrow\ \text{e}^{\tan^{-1}\text{x}}\frac{\text{dy}}{\text{dx}}+\frac{\text{y}\text{e}^{\tan^{-1}\text{x}}}{1+\text{x}^2}=\text{e}^{\tan^{-1}\text{x}}\frac{\text{e}^{\tan^{-1}\text{x}}}{1+\text{x}^2}$
Integrating both sides with respect to x, we get
$\text{y}\text{e}^{\tan^{-1}\text{x}}=\int\frac{\text{e}^{2\tan^{-1}\text{x}}}{1+\text{x}^2}\text{dx + C}$
$\Rightarrow\ \text{y}\text{e}^{\tan^{-1}\text{x}}=\text{I + C}\ \dots(2)$
Here,
$\text{I}=\int\frac{\text{e}^{2\tan^{-1}\text{x}}}{1+\text{x}^2}\text{dx}$
Putting $\tan^{-1}\text{x}=\text{t},$ we get
$\frac{1}{1+\text{x}^2}\text{dx}=\text{dt}$
$\therefore\ \text{I}=\int\text{e}^{2\text{t}}\text{dt}$
$=\frac{\text{e}^{2\text{t}}}{2}$
$=\frac{\text{e}^{2\tan^{-1}\text{x}}}{2}$
Putting the value of I in (2), we get
$\text{y}\text{e}^{\tan^{-1}\text{x}}=\frac{{\text{e}^{2\tan^{-1}\text{x}}}}2+\text{C}$
$\Rightarrow\ 2\text{y}\text{e}^{\tan^{-1}\text{x}}=\text{e}^{2\tan^{-1}\text{x}}+2\text{C}$
$\Rightarrow\ 2\text{y}\text{e}^{\tan^{-1}\text{x}}=\text{e}^{2\tan^{-1}\text{x}}+\text{K}$ (where K = 2C)
Hence, $2\text{y}\text{e}^{\tan^{-1}\text{x}}=\text{e}^{2\tan^{-1}\text{x}}+\text{K}$ 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 :

$\int \frac{7 x+3}{\sqrt{3+2 x-x^2}} d x$

→
Five defective mangoes are acciedently mixed with 15 good ones. Four mangoes are drawn at random from this lot. Find the probability distribution of the number of defective mangoes.
→
Solve the following differential equation:
$(\text{x}^2-1)\frac{\text{dy}}{\text{dx}}+2(\text{x}+2)\text{y}=2(\text{x}+1)$
→
Find $\frac{\text{dy}}{\text{dx}},$ when
$\text{x}=\frac{3\text{at}}{1+\text{t}^2}\text{ and y}=\frac{3\text{at}^2}{1+\text{t}^2}$
→
Solve the following differential equations:

$y \log y=\left(\log y^2-x\right) \frac{d y}{d x}$

→
Solve the following differential equations:$\frac{\text{dy}}{\text{dx}}=2\text{e}^{2\text{x}}\text{y}^2,\text{y}(0)=-1$
→
If $\text{x}=\text{a}\sin\text{t}-\text{b}\cos\text{t},\text{y}=\text{a}\cos\text{t}+\text{b}\sin\text{t},$ Prove that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{x}^2+\text{y}^2}{\text{y}^2}$
→
Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane $\vec{\text{r}}\cdot(\hat{\text{i}}+2\hat{\text{j}}-5\hat{\text{k}})+9=0.$
→
A beam is supported at the two ends and is uniformly loaded. The bending moment M at a distance x from one end is given by
$\text{M}=\frac{\text{WL}}{2}\text{x}-\frac{\text{W}}{2}\text{x}^{2}$
Find the point at which M is maximum in each case.
→
Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theorem.
$f(x) = x^3- 5x^2 - 3x$ on $[1, 3]$
→