$\begin{array}{1 1} y=x-4-2e^x \\ y=x-4+2e^x \\ y=x+4-2e^x \\ y=-x+4-2e^x \end{array}$

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

- Slope of the curve is $\large\frac{dy}{dx}$
- To solve a first order linear differential equation of the form $\large\frac{dy}{dx}$$ + Py = Q$
- (i) First write the equation in the form $\large\frac{dy}{dx}$$ + Py = Q$
- (ii) Find the integral Factor (I.F) =$ e^{\int Pdx}$
- Write the solution as y(I.F) = integration of Q. (I.F) dx + C

Step 1:

According to the given information, let F(x,y) be the curve and dy/dx be the slope to the curve.

Hence $\large\frac{dy}{dx }$$ + 5 = x + y$

Let us rewrite the equation as $\large\frac{dy}{dx}$$ -y = x + 5$

The equation is of the form $\large\frac{dy}{dx}$$ + Py = Q$

Here $P = -1$ and $Q = x + 5$

Step 2:

To find the integrating factor

$\int -dx=-x$

Hence $I.F = e^{-x}$

The required solution is $ye^{-x} = \int(x+5).e^{-x}.dx + C$

$\int(x+5) e^{-x }dx$ can be done by parts.

Let $u = x+5$, hence $du = dx$ and $dv = e^{-x}dx ;v = -e^{-x}$

$\int(x+5)e^{-x} = (x+5)(-e^{-x}) -\int(-e^{-x})dx$

$\qquad\qquad\;\;=-(x+5)e^{-x}+\int e^{-x}dx$

$\qquad\qquad\;\;=-(x+5)e^{-x}-e^{-x}+C$

Hence the required solution is $ye^{-x }= -(x+5)e^{-x} - e^{-x} + C$

$ye^-x = (5-x)e^{-x} - e^{-x} + C $

$ye^-x = (4-x)e^{-x} + C$

Step 3:

To evaluate $C$, let us substitute the values of $x$ and $y$ in the above equation.

$2e^0 = (4-0)e^0 + C$

$C = - 2$

Substituting for $C$ we get,

$ye^{-x} = (4 - x) e^{-x} - 2$

dividing throughout by $e^{-x}$ we get,

$y = (4-x) - \large\frac{2}{e^{-x}}$

$y = x - 4 - 2e^x$ is the required solution.

Ask Question

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...