Download App

DIFFERENTIAL EQUATIONS — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSDIFFERENTIAL EQUATIONS5 Marks
Question
Solve the following differential equation:
$\text{x}\frac{\text{dy}}{\text{dx}}+2\text{y}=\text{x}\cos\text{x}$
✓

Answer

Here, $\frac{\text{dy}}{\text{dx}}+\frac{2\text{y}}{\text{x}}=\cos\text{x}$ It is a linear differential equation. Comparing it with, $\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$ $\text{P}=\frac{2}{\text{x}},\text{Q}=\cos\text{x}$ I.F. $=\text{e}^{\int\text{Pdx}}$ $=\text{e}^{2\int\frac{1}{\text{x}}\text{dx}}$ $=\text{e}^{2\log|\text{x}|}$ $=\text{x}^2$Solution of the equation is given by,
$\text{y}\times(\text{I.F.}=\int\text{Q}\times(\text{I.F.})\text{dx + C}$ $\text{y}(\text{x}^2)=\int\cos\text{x}(\text{x}^2)\text{dx + C}$ $\text{yx}^2=\int\text{x}^2\cos\text{xdx + C}$ $=\text{x}^2\int\cos\text{x}-\int(2\text{x}\times\int\cos\text{xdx})\text{dx + C}$ Usind integration by parts $\text{yx}^2=\text{x}^2\sin\text{x}-\int2\text{x}\sin\text{xdx + C}$ $=\text{x}^2\sin\text{x}-2\big[\text{x}\times\int\sin\text{xdx}-\int(1\times\int\sin\text{xdx})\text{dx}\big]+\text{C}$ $=\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\int\cos\text{xdx + C}$ $\text{yx}^2=\text{x}^2\sin\text{x}+2\text{x}\cos\text{x}-2\sin\text{x + C}$ $\text{y}=\sin\text{x}+\frac{2}{\text{x}}\cos\text{x}-\frac{2}{\text{x}^2}\sin\text{x}+\frac{\text{C}}{\text{x}^2}$

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→Rajasthan Board STD 12 Science question papers→Rajasthan Board question paper generator→

Similar questions

If $\text{A}=\begin{bmatrix}2&-3&-5\\-1&4&5\\1&-3&-4\end{bmatrix}$ and $\text{B}=\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix},$ show that AB = A and BA = B.
→
Without expanding, prove that:
$\begin{vmatrix}\text{a}&\text{b}&\text{c}\\\text{x}&\text{y}&\text{z}\\\text{p}&\text{q}&\text{r}\end{vmatrix}=\begin{vmatrix}\text{x}&\text{y}&\text{z}\\\text{p}&\text{q}&\text{r}\\\text{a}&\text{b}&\text{c}\end{vmatrix}=\begin{vmatrix}\text{y}&\text{b}&\text{q}\\\text{x}&\text{a}&\text{p}\\\text{z}&\text{c}&\text{r}\end{vmatrix}$
→
Evaluate the following integrals:
$\int(2\text{x}-5)\sqrt{2+3\text{x}-\text{x}^2}\text{dx}$
→
It is given that the Rolle's theorem holds for the function $f(x) = x^3 + bx^2 + cx,$ $\text{x}\in[1,2]$ at the point $\text{x}=\frac{4}{3},$ the values of $b$ and $c.$
→
Find the inverse of the following matrices by using elementry row transformation:$\begin{bmatrix}7 & 1 \\4 & -3 \end{bmatrix}$
→
Show that the four points A, B, C, D with position vectors $\vec{\text{a}},\ \vec{\text{b}},\ \vec{\text{c}},\ \vec{\text{d}}$ respectively such that $3\vec{\text{a}}-2\vec{\text{b}}+5\vec{\text{c}}-6\vec{\text{d}}=0$, are coplanar. Also, find the position vector of the point of intersection of the line segments AC and BD.
→
Solve the following differential equation:
$\frac{\text{dy}}{\text{dx}}+1=\text{e}^{\text{x + y}}$
→
Evaluate the following integrals:
$\int\frac{\text{e}^{\text{x}}(\text{x}-4)}{(\text{x}-2)^3}\text{dx}$
→
Evaluate the following integrals:
$\int^\limits{\pi}_0\sin^3\text{x}(1+2\cos\text{x})(1+\cos\text{x})^2\text{ dx}$
→
Find the area, lying above $x-$axis and included between the circle $x^2 + y^2 = 8x$ and the parabola $y^2 = 4x.$
→