Answer to Question #213431 in Differential Equations for joshua

Question

1.given the system express y as a function of t

1. 3dx/dt+2dy/dt-x+y=t-1
2. dx/dt+dy/dt-x=t+2

2.what do you understand by ordinary paint of the equation

a₀(x)d²y/dx²+a₁(x)dy/dx+a₂(x)y=0? hence using taylor series expansion find a series solution in powers of x for the equation(2x³-3)d²y/dx²-2xdy/dx+y=0 y(0)=-1

y¹(0)=5

Answer 1

1)

Given the system let us express $y$ as a function of $t .$

$i)$ $3\dfrac{dx}{dt}+2\dfrac{dy}{dt}-x+y=t-1$

$ii)\dfrac{dx}{dt}+\dfrac{dy}{dt}-x=t+2$

Multiplying equation by $\times 3$ , we get -

And the adding $i)\ and \ ii)\ ,$ we get -

$=\dfrac{-dy}{dt}+2x+y=-2t-7$

It gives values of $x$ as -

$x=\dfrac{1}{2}\dfrac{dy}{dt}-\dfrac{1}{2}y-t-\dfrac{7}{2}$

Now , differentiating with respect to t , we get -

$=\dfrac{dx}{dt}=$ $\dfrac{1}{2}\dfrac{d^{2}y}{dt^{2}}-\dfrac{1}{2}\dfrac{dy}{dt}-1$ .

We get the equation as -

$=\dfrac{1}{2}\dfrac{d^{2}y}{dt^{2}}-\dfrac{1}{2}\dfrac{dy}{dt}-1+\dfrac{dy}{dt}-(\dfrac{1}{2}\dfrac{dy}{dt}-\dfrac{1}{2}y-t-\dfrac{7}{2})=t+2$

Which is equivalent to -

$=\dfrac{1}{2}\dfrac{d^{2}y}{dt^{2}}+\dfrac{1}{2}y=\dfrac{-1}{2}$

And hence the equation becomes -

$=\dfrac{d^{2}y}{dt^{2}}+y=-1$

Auxiliary equation becomes -

$=m^{2}+1=0$

It's root's becomes as -

$m=+i,-i$ ,

It's follows that solution is of form -

$=y(t)=C_1cost+C_2sint +y_p,$ where $y_p=a$

It's follows that -

$\dfrac{d^{2}y_p}{dt^{2}}=0$ , hence $a=-1.$ Therefore the solution is the following -

$y(t)=C_1cost +C_2sint-1$

$2)$

Ordinary point of equation is defined as -

A point a is called ordinary point of differential equation when function $p_{1}(x)\ and \ p_0(x)$ are analytical at $x=a$ ,

$=a_0(x)\dfrac{d^{2}y}{dx^{2}}+a_1(x)\dfrac{dy}{dx}+a_2(x)y=0$

For , the above equation the ordinary point is called as -

A point $x=x_0$ is called ordinary point of differential equation if $a_0(x)\neq0$ at $x=x_0$ . The point $x=x_0$ is called that ordinary point .

$=(2x^{3}-3)\dfrac{d^{2}y}{dx^{2}}-2x\dfrac{dy}{dx}+y=0$ $y(0)=-1$

$y^{'}(1)=5$

$Now$ , we know that taylor's series expansion is expressed as -

$y(x)=$ $y(0)+y^{'}(0)(x-0)+\dfrac{1}{2!}y^{''}(0)(x-0)^{2}+....$

$=y^{''}(x)=\dfrac{2xy^{'}(x)}{2x^{3}-3}-\dfrac{y}{2x^{3}-3}$

Now putting , $x=0,$ we get -

$y^{''}(0)=0-\dfrac{y(0)}{-3}$

$y^{''}(0)=\dfrac{-1}{3}$

$y(x)=-1+5x+\dfrac{1}{2!}\dfrac{-1}{3}x^{2}+.........$

This is taylor's series of given equation .