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Chapter 12: Q86SE (page 541)

In Problems 27鈥�30 use (12) of Section 1.1 to verify that the indicated function is a solution of the given differential equation. Assume an appropriate interval I of definition of each solution.

\({x^2}y'' + \left( {{x^2} - x} \right)y' + (1 - x)y = 0;\;\;\;y = x\int_1^x {\frac{{{e^{ - t}}}}{t}} dt\)

Short Answer

Expert verified

The indicated function is a solution of the differential function.

Step by step solution

01

Simplify the given differential equation.

Let the given differential equation be\({\rm{y = x}}\int_{\rm{1}}^{\rm{x}} {\frac{{{{\rm{e}}^{{\rm{ - t}}}}}}{{\rm{t}}}} {\rm{dt}}\).

Divide each side of the equation by\({\rm{x}}\).

\(\frac{{\rm{y}}}{{\rm{x}}}{\rm{ = }}\int_{\rm{1}}^{\rm{x}} {\frac{{{{\rm{e}}^{{\rm{ - t}}}}}}{{\rm{t}}}} {\rm{\;dt}}\)

02

Determine the first derivative.

Take differential on both sides of the equation.

Multiply\({\rm{x}}\)on both sides of the equation.

\(\begin{array}{c}\frac{{{\rm{y'x - y}}}}{{\rm{x}}}{\rm{ = }}{{\rm{e}}^{{\rm{ - x}}}}\\{\rm{y' - }}\frac{{\rm{y}}}{{\rm{x}}}{\rm{ = }}{{\rm{e}}^{{\rm{ - x}}}}\end{array}\)

03

Determine the solution of the indicated function.

Take differential on both sides of the equation.

\(y'' - \frac{{y'x - y}}{{{x^2}}} = - {e^{ - x}}\)

Add both the differentials to get,

\(\begin{array}{c}{\rm{y' - }}\frac{{\rm{y}}}{{\rm{x}}}{\rm{ + y'' - }}\frac{{{\rm{y'x - y}}}}{{{{\rm{x}}^{\rm{2}}}}}{\rm{ = }}{{\rm{e}}^{{\rm{ - x}}}}{\rm{ - }}{{\rm{e}}^{{\rm{ - x}}}}\\\frac{{{{\rm{x}}^{\rm{2}}}{\rm{y'' + }}\left( {{{\rm{x}}^{\rm{2}}}{\rm{ - x}}} \right){\rm{y' + (1 - x)y}}}}{{{{\rm{x}}^{\rm{2}}}}}{\rm{ = 0}}\\{{\rm{x}}^{\rm{2}}}{\rm{y'' + }}\left( {{{\rm{x}}^{\rm{2}}}{\rm{ - x}}} \right){\rm{y' + (1 - x)y = 0}}\end{array}\)

Hence, the indicated function is a solution of the differential function.

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Most popular questions from this chapter

Rotating Fluid As shown in Figure 1.3.24(a), a right-circular cylinder partially filled with fluid is rotated with a constant angular velocity v about a vertical \(y - \)axis through its center. The rotating fluid forms a surface of revolution S. To identify S, we first establish a coordinate system consisting of a vertical plane determined by the \(y - \)axis and an \(x - \)axis drawn perpendicular to the \(y - \)axis such that the point of intersection of the axes (the origin) is located at the lowest point on the surface S. We then seek a function \(y = f(x)\) that represents the curve C of intersection of the surface S and the vertical coordinate plane. Let the point \(P(x,y)\) denote the position of a particle of the rotating fluid of mass m in the coordinate plane. See Figure 1.3.24(b).

(a) At \(P\) there is a reaction force of magnitude F due to the other particles of the fluid which is normal to the surface \(S\). By Newton鈥檚 second law the magnitude of the net force acting on the particle is \(m{\omega ^2}x\). What is this force? Use Figure 1.3.24(b) to discuss the nature and origin of the equations

\(F\cos \theta = mg,F\sin \theta = m{\omega ^2}x.\)

(b) Use part (a) to find a first-order differential equation that denes the function \(y = f(x)\).

In Problems 7鈥�12 match each of the given differential equations with one or more of these solutions:

(a) \(y = 0\), (b) \(y = 2\), (c) \(y = 2x\), (d) \(y = 2{x^2}\)

\(y'' + 9y = 18\)

In Problems \(11 - 14\) verify that the indicated function is an explicit solution of the given differential equation. Assume an appropriate interval I of definition for each solution.

\(y'' - 6y' + 13y = 0; y = - (cos x) ln(sec x + tan x)\).

In Problems \(3\) and \(4\) Fill in the blank and then write this result as a linear second-order differential equation that is free of the symbols andand has the form. The symbols,, andrepresent constants. \({c_1}\)

\(\frac{{{d^2}}}{{d{x^2}}}({c_1}cosh kx + {c_2}sinh kx) = \_\_\_\_\_\)

In Problems \(1 - 8\) state the order of the given ordinary differential equation. Determine whether the equation is linear or nonlinear by matching it with \((6)\).

\(\frac{{{d^2}u}}{{d{r^2}}} + \frac{{du}}{{dr}} + u = cos(r + u)\)
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