Differential Equations Test - 25

Given  $$\displaystyle \left ( e^{y}+1 \right )\cos x dx+e^{y}\sin x dy=0$$

As $$m_{1}$$, $$m_{2}$$, $$m_{3}$$ are roots of $$m^{3}-9m^{2}+23m-15=0$$
$$\Rightarrow $$   $$\left ( m-1 \right )\left ( m-3 \right )\left ( m-5 \right )=0$$
$$\Rightarrow $$   $$m_{1}=1$$, $$m_{2}=3$$, $$m_{3}=5$$
$$\therefore $$   Required solution is $$y=c_{1}e^{x}+c_{2}e^{3x}+c_{3}e^{5x}$$          (i)
On Differentiating w.r to x both sides we get
$$\Rightarrow $$   $${y}'=c_{1}e^{x}+3c_{2}e^{3x}+5c_{3}e^{5x}$$          (ii)
(using (ii) -(i))
$$\Rightarrow $$   $${y}'-y=2c_{2}e^{3x}+4c_{3}e^{5x}$$          (iii)
Again differentiating w.r to x both sides
$$\Rightarrow $$   $${y}''-{y}'=6c_{2}e^{3x}+20c_{3}e^{5x}$$          (iv)
using (iv)-3(iii) we get
     $${y}''-4{y}'+3y=8c_{3}e^{5x}$$          (v)
$$\Rightarrow $$   $${y}'''-4{y}''+3{y}'=8\times 5c_{3}e^{5x}$$          (vi)
using (vi)-5(v) we get
     $${y}'''-9{y}''+23{y}'-15y=0$$
or $$\displaystyle \frac{d^{3}y}{dx^{3}}-9\frac{d^{2}y}{dx^{2}}+23\frac{dy}{dx}-15y=0$$
Hence order of differential equation is 3.

Given, the parabola is $$(x-h)^2=4a(y-k)$$
Since, there are 3 unknown constants.
Therefore, order of differential equation will be 3

Given  $$\displaystyle \left ( e^{x}+1 \right )y dy=\left ( y+1 \right )e^{x}dx$$

Given $${f}''\left ( x \right )=6\left ( x-1 \right )$$
$$\Rightarrow $$   $$\displaystyle {f}'\left ( x \right )=\frac{6\left ( x-1 \right )^{2}}{2}+c$$
$$\Rightarrow $$   $$3=3+c$$          $$\left \{ \because f\left ( x \right )=y=3x+5, {f}'\left ( x \right )=3\forall x\epsilon R \right \}$$
$$\Rightarrow $$   $$c=0$$
so $${f}'\left ( x \right )=3\left ( x-1 \right )^{2}$$
$$\Rightarrow $$   $$f\left ( x \right )=\left ( x-1 \right )^{3}+c_{1}$$ as curve passes through (2, 1)
$$\Rightarrow $$   $$1=\left ( 2-1 \right )^{3}+c_{1}$$
$$\Rightarrow $$   $$c_{1}=0$$
$$\therefore $$   $$f\left ( x \right )=\left ( x-1 \right )^{3}$$

Given,  $$\displaystyle \left ( 1-x \right )dy-\left ( 3+y \right )dx=0$$
 $$\Rightarrow \displaystyle \frac{dy}{3+y}=\frac{dx}{1-x}$$
or $$\displaystyle \int \frac{dy}{3+y}=\int \frac{dx}{1-x}+c$$
or $$\displaystyle \log \left ( 3+y \right )=-\log \left ( 1+x \right )+c,$$
or $$\displaystyle \log \left ( 3+y \right )+\log \left ( 1-x \right )=c,$$
or $$\displaystyle \log \left ( 3+y \right )\left ( 1-x \right )=c,or \log k.$$
$$\displaystyle \therefore \left ( 3+y \right )\left ( 1-x \right )=k$$ is the required solution. c being arbitrary constant, can be put equal to
$$\displaystyle \log k$$ for the sake of simplification.

$$\dfrac { d^{ 2 }y }{ dx^{ 2 } } =e^{ -2x }$$
Integrating both sides w.r.t. x
$$\dfrac { dy }{ dx } =\cfrac { e^{ -2x } }{ -2 } +c$$
Again integrating w.r.t. x
$$y=\dfrac { e^{ -2x } }{ 4 } +cx+d$$

Given $$\displaystyle \left ( 1-x^{2} \right )\left ( 1-y \right )dx=xy\left ( 1+y \right )dy.$$

$$\displaystyle ydx-x dy=xy\:dx$$

Given, $$\displaystyle \left ( xy^{2}+x \right )dx+\left ( yx^{2}+y \right )dy=0.$$