Question
Solve the following differential equation:
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=(\text{x}-1)\text{e}^{\text{x}}$
$\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=(\text{x}-1)\text{e}^{\text{x}}$
✓
Answer
Here, $\text{x}\frac{\text{dy}}{\text{dx}}-\text{y}=(\text{x}-1)\text{e}^{\text{x}}$
$\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}}=\Big(\frac{\text{x}-1}{\text{x}}\Big)\text{e}^{\text{x}}$
It is a linear differential equation. comparing the equation by,
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
$\text{P}=-\frac{1}{\text{x}},\text{Q}=\Big(\frac{\text{x}-1}{\text{x}}\Big)\text{e}^{\text{x}}$
I.F. $=\text{e}^{\int\text{Pdx}}=\text{e}^{-\int\frac{1}{\text{x}}\text{dx}}$
$\text{e}^{-\log|\text{x}|}=\frac{1}{\text{x}},\text{x}>0$
Solution of the equation is given by,
$\text{y}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{dx + C}$
$\text{y}\Big(\frac{1}{\text{x}}\Big)=\int\Big(\frac{\text{x}-1}{\text{x}}\Big)\text{e}^{\text{x}}\text{dx + C}$
$\frac{\text{y}}{\text{x}}=\int\Big(\frac{1}{\text{x}}-\frac{1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{dx + C}$
$\frac{\text{y}}{\text{x}}=\frac{1}{\text{x}}\text{e}^{\text{x}}+\text{C}$
Since $\int[\text{f(x)}+\text{f}'(\text{x})]\text{e}^{\text{x}}\text{dx}=\text{f(x)}\text{e}^{\text{x}}+\text{C}$
$\text{y}=\text{e}^{\text{x}}+\text{Cx},\text{x}>0$
$\frac{\text{dy}}{\text{dx}}-\frac{\text{y}}{\text{x}}=\Big(\frac{\text{x}-1}{\text{x}}\Big)\text{e}^{\text{x}}$
It is a linear differential equation. comparing the equation by,
$\frac{\text{dy}}{\text{dx}}+\text{Py}=\text{Q}$
$\text{P}=-\frac{1}{\text{x}},\text{Q}=\Big(\frac{\text{x}-1}{\text{x}}\Big)\text{e}^{\text{x}}$
I.F. $=\text{e}^{\int\text{Pdx}}=\text{e}^{-\int\frac{1}{\text{x}}\text{dx}}$
$\text{e}^{-\log|\text{x}|}=\frac{1}{\text{x}},\text{x}>0$
Solution of the equation is given by,
$\text{y}\times(\text{I.F.})=\int\text{Q}\times(\text{I.F.})\text{dx + C}$
$\text{y}\Big(\frac{1}{\text{x}}\Big)=\int\Big(\frac{\text{x}-1}{\text{x}}\Big)\text{e}^{\text{x}}\text{dx + C}$
$\frac{\text{y}}{\text{x}}=\int\Big(\frac{1}{\text{x}}-\frac{1}{\text{x}^2}\Big)\text{e}^{\text{x}}\text{dx + C}$
$\frac{\text{y}}{\text{x}}=\frac{1}{\text{x}}\text{e}^{\text{x}}+\text{C}$
Since $\int[\text{f(x)}+\text{f}'(\text{x})]\text{e}^{\text{x}}\text{dx}=\text{f(x)}\text{e}^{\text{x}}+\text{C}$
$\text{y}=\text{e}^{\text{x}}+\text{Cx},\text{x}>0$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
At what points on the curve x2 + y2 - 2x - 4y + 1 = 0, the tangents are parallel to the y-axis?
Find the angle between the lines whose direction ratios are proportional to a, b, c and b - c, c - a, a - b.
Solve the following systems of linear equations by cramer's rule:
x - 2y = 4,
-3x + 5y = -7
If $\text{f(x)}=\begin{cases}\text{ax}^2-\text{b}, & \text{if |x|}<1\\\frac{1}{|\text{x}|}, & \text{if |x|}\geq1\end{cases}$ is differentiable at x = 1, find a, b.
→Evaluate the following intregals:
$\int\frac{1}{(\sin\text{x}-2\cos\text{x})(2\sin\text{x}-\cos\text{x})}\ \text{dx}$
→$\int\frac{1}{(\sin\text{x}-2\cos\text{x})(2\sin\text{x}-\cos\text{x})}\ \text{dx}$
A dietician wishes to mix two types of foods in such a way that the vitamin contents of the mixture contains at least 8 units of vitamin A and 10 units of vitamin C. Food I contains 2 units/kg of vitamin A and 1 unit/kg of vitamin C while Food II contain 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs 
5 per kg to purchase Food I and 
7 per kg to purchase Food II. Determine the minimum cost of such a mixture. Formulate the above as a LPP and solve it graphically.
5 per kg to purchase Food I and 7 per kg to purchase Food II. Determine the minimum cost of such a mixture. Formulate the above as a LPP and solve it graphically. If $\text{A}=\begin{bmatrix}0&1\\1&1\end{bmatrix}$ and $\text{B}=\begin{bmatrix}0&-1\\1&0\end{bmatrix},$ then show that $(\text{A}+\text{B})(\text{A}-\text{B})\neq\text{A}^2-\text{B}^2.$
→Consider $\text{f : R} - \left\{-\frac{4}{3}\right\} \rightarrow \text{R} - \left\{\frac{4}{3}\right\} \text{given by f(x)} = \frac{\text{4x + 3}}{\text{3x + 4}}.$ Show that f is bijective. Find the inverse of f and hence find f –1(0) and x such that f –1(x) = 2.
→$\text{A(adj. A) = (adj.A)A} | \text{A} = | \text{I}_{3}.$
If a, b and c are all non-zero and $ \begin{vmatrix} \text{1 + a} & 1 & 1 \\ 1 & \text{1 + b} & 1 \\ 1 & 1 & \text{1 + c} \end{vmatrix} = 0,$ then prove that $\frac{1}{\text{a}} + \frac{1}{\text{b}} + \frac{1}{\text{c}} 1 = 0.$
→If a, b and c are all non-zero and $ \begin{vmatrix} \text{1 + a} & 1 & 1 \\ 1 & \text{1 + b} & 1 \\ 1 & 1 & \text{1 + c} \end{vmatrix} = 0,$ then prove that $\frac{1}{\text{a}} + \frac{1}{\text{b}} + \frac{1}{\text{c}} 1 = 0.$
If $(\cos\text{x})^\text{y}=(\cos\text{y})^\text{x},$ find $\frac{\text{dy}}{\text{dx}}$
→