Download App

Differential Equations — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsDifferential Equations5 Marks
Question
Solve the following initial value problems:
$\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}=2\text{x}+\text{x}^2\tan\text{x},\text{ y}(0)=1$
✓

Answer

We have,
$\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}=2\text{x}+\text{x}^2\tan\text{x}\ ...(1)$
Clearly, it is a linear differential equation of the form
$\frac{\text{dy}}{\text{dx}}+\text{Px}=\text{Q}$
Where $\text{P}=\tan\text{x}$ and $\text{Q}=\text{x}^2\cot\text{x}+2\text{x}$
$\therefore\text{ I.F.}=\text{e}^{\int\text{Pdx}}$
$=\text{e}^{\int\tan\text{x dx}}$
$=\text{e}^{\log|\sec\text{x}|}$
$=\sec\text{x}$
Multiplying both sides of (1) by $\text{I.F.}=\sec\text{x},$ we get
$\sec\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}\Big)=\sec\text{x}(\text{x}^2\tan\text{x}+2\text{x})$
$\Rightarrow\sec\text{x}\Big(\frac{\text{dy}}{\text{dx}}+\text{y}\tan\text{x}\Big)=\text{x}^2\tan\text{x }\sec\text{x}+2\text{x}\sec\text{x}$
Integrating both sides with respect to x, we get
$\text{y}\sec\text{x}=\int\text{x}^2\tan\text{x }\sec\text{x dx}+2\int\text{x}\sec\text{x dx}\\+2\int\text{x}\sec\text{x dx} +\text{C}$
$\Rightarrow\text{y}\sec\text{x}=\int\text{x}^2\tan\text{x }\sec\text{x dx}+2\sec\text{x}\int\text{x dx}\\-2\int\Big[\frac{\text{d}}{\text{dx}}(\sec\text{x})\int\text{x dx}\Big]\text{dx}+\text{C}$
$\Rightarrow\text{y}\sec\text{x}=\int\text{x}^2\tan\text{x }\sec\text{x dx}+\text{x}^2\sec\text{x}\\-\int\text{x}^2\tan\text{x }\sec\text{x dx} +\text{C}$
$\Rightarrow\text{y}\sec\text{x}=\text{x}^2\sec\text{x}+\text{C}$
$\Rightarrow\text{y}=\text{x}^2+\text{C}\cos\text{x}\ ...(2)$
Now,
$\text{y}(0)=1$
$\therefore\ 1=0+\text{C}\cos0$
$\Rightarrow\text{C}=1$
Putting the value of C in (2), we get
$\text{y}=\text{x}^2+\cos\text{x}$
Hence, $\text{y}=\text{x}^2+\cos\text{x}$ 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

Solve the following differential equations:
$\frac{\text{dy}}{\text{dx}}=1+\text{x}^2+\text{y}^2+\text{x}^2\text{y}^2,\text{y}(0)=1$
→
Solve the following differential equation
$\text{x}(\text{x}^{2} - 1)\frac{\text{dy}}{\text{dx}} = 1, \text{y}(2) = 0$
→
If $\text{x}=\text{a}\cos\theta,\text{y}=\text{b}\sin\theta$ Show that $\frac{\text{d}^2\text{y}}{\text{dx}^2}=-\frac{\text{b}^4}{\text{a}^2\text{y}^3}$
→
Let $\text{A}=\begin{bmatrix}3 & 2&7 \\1 & 4&3\\-2&5&8 \end{bmatrix}.$ Find matrices X and Y such that X + Y = A, where X is a symmetric and Y is a skew-symmetric matrix.
→
Evaluate the following integrals:
$\int\frac{\text{x}^2}{\text{x}^4+\text{x}^2-2}\ \text{dx}$
→
There are three categories of students in a class of 60 students:
A : Very hardworking
B : Regular but not so hardworking
C : Careless and irregular 10 students are in category A, 30 in category B and the rest in category C.
It is found that the probability of students of category A, unable to get good marks in the final year examination is 0.002, of category B it is 0.02 and of category C, this probability is 0.20. A student selected at random was found to be one who could not get good marks in the examination. Find the probability that this student is category C.
→
Find the matrix X satisfying the matrix equation.$\text{X}\begin{bmatrix}5 & 3 \\-1 & -2 \end{bmatrix}\begin{bmatrix}14 & 7 \\7 & 7 \end{bmatrix}$
→
Evaluate : $\int \frac{\cos \theta}{\cos 3 \theta} \cdot d \theta$
→
In the following, find the value of the constant k so that the given function is continuous at the indicated point:
$\text{f(x)}=\begin{cases}(\text{x}-1)\tan\frac{\pi\text{x}}{2},&\text{if}\text{ x}\neq1\\\text{k},&\text{if}\text{ x}=1\end{cases}\text{at x} = 1$
→
Find the slopes of the tangent and the normal to the following curves at the indicated points:
$\text{x}=\text{a}(\theta-\sin\theta),\text{y}=\text{a}(1-\cos\theta)\ \text{at}\ \theta=\frac{-\pi}{2}$
→