i need help with the following question attached.

thanks

2. The equation is...

$y^{'}-2\cdot y=-5\cdot x$ (1)

... and its caratteristic equation...

$a-2=0$

... has solution $a=2$. So the general integral of 'incomplete equation' is...

$y_{i}(x)= c\cdot e^{2x}$ (2)

It is aesy to verify that a particular integral of (1) is...

$y_{p}(x)= -\frac{5}{2}\cdot x -\frac{5}{4}$ (3)

... so that the general integral of (1) is...

$y(x)= c\cdot e^{2x} -\frac{5}{2}\cdot x -\frac{5}{4}$ (4)

If we insert the 'initial condition' we obtain...

$y(x)= \frac{17}{4}\cdot e^{2x} -\frac{5}{2}\cdot x -\frac{5}{4}$ (5)

Kind regards

$\chi$ $\sigma$

3. Hello, cooltowns!

It's fairly straight forward . . . Where is your difficulty?

Solve: . $\frac{dy}{dx} - 2y \:=\:5x,\quad y(0) = 3$

Integrating factor: . $I \:=\:e^{\int \text{-}2\,dx} \:=\:e^{\text{-}2x}$

We have: . $e^{-2x}\frac{dy}{dx} - 2e^{-2x} y \;=\;5xe^{-2x}$

. . . . . . . $\frac{d}{dx}\left(e^{2x}y\right) \:=\:5xe^{-2x}$

Integrate: . $e^{2x}y \;=\;5\!\!\int\!\! xe^{-2x}dx$

. . By parts: . $\begin{array}{ccccccc}u &=& x & & dv &=& e^{-2x}dx \\ \\[-4mm] du &=& dx & & v &=&-\frac{1}{2}e^{-2x} \end{array}$

We have: . $e^{-2x}y \;=\;5\bigg[\text{-}\tfrac{1}{2}xe^{-2x} + \tfrac{1}{2}\!\!\int e^{-2x}dx\bigg] \;=\;5\bigg[\text{-}\tfrac{1}{2}xe^{-2x} -\tfrac{1}{4}e^{-2x}\bigg] + C$

. . . . . . . $e^{-2x}y \;=\;-\tfrac{5}{2}xe^{-2x} - \tfrac{5}{4}e^{-2x} + C$

Multiply by $e^{2x}\!:\quad y \;=\;-\tfrac{5}{2}x - \tfrac{5}{4} + Ce^{2x}$

Since $y(0)=3\!:\quad3 \;=\;-\tfrac{5}{2}(0) - \tfrac{5}{4} + Ce^0 \quad\Rightarrow\quad C \:=\:\frac{17}{4}$

Therefore: . $y \;=\;-\frac{5}{2}x - \frac{5}{4} + \frac{17}{4}e^{2x}$

4. i haven't done this type of differential equation before i didn't know you have use intergrating factors.