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# 1.2E: Basic Concepts (Exercises)

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## Exercises

[exer:1.2.1} Find the order of the equation.

1. $${d^2y\over dx^2}+2 {dy\over dx}\ {d^3y\over dx^3}+x=0$$
2. $$y''-3y'+2y=x^7$$
3. $$y'-y^7=0$$
4. $$y''y-(y')^2=2$$

[exer:1.2.2} Verify that the function is a solution of the differential equation on some interval, for any choice of the arbitrary constants appearing in the function.

$$y=ce^{2x}; \quad y'=2y$$

$$y= {x^2\over3} +{c\over x}; \quad xy'+y=x^2$$

$$y= {1\over2}+ce^{-x^2}; \quad y'+2xy=x$$

$$y=(1+ce^{-x^2/2}); (1-ce^{-x^2/2})^{-1} \quad 2y'+x(y^2-1)=0$$

$$y= {\tan\left( {x^3\over3}+c\right)}; \quad y'=x^2(1+y^2)$$

$$y=(c_1+c_2x)e^x+\sin x+x^2; \quad y''-2y'+y=-2 \cos x+x^2-4x+2$$

$$y=c_1e^x+c_2x+ {2\over x}; \quad (1-x)y''+xy'- y=4(1-x-x^2)x^{-3}$$

$$y=x^{-1/2}(c_1\sin x+c_2 \cos x)+4x+8$$; $$x^2y''+xy'+ {\left(x^2-{1\over4}\right)}y=4x^3+8x^2+3x-2$$

[exer:1.2.3} Find all solutions of the equation.

(a)$$y'=-x$$ & (b)$$y'=-x \sin x$$
(c)$$y'=x \ln x$$ & (d)$$y''=x \cos x$$
(e)$$y''=2xe^x$$ & (f)$$y''=2x+\sin x+e^x$$
(g)$$y'''=-\cos x$$ & (h)$$y'''=-x^2+e^x$$
(i)$$y'''=7e^{4x}$$&

[exer:1.2.4} Solve the initial value problem.

$$y'=-xe^x, \quad y(0)=1$$

$${y'=x \sin x^2, \quad y\left({\sqrt{\pi\over2}}\right)=1}$$

$$y'=\tan x, \quad y(\pi/4)=3$$

$$y''=x^4, \quad y(2)=-1, \quad y'(2)=-1$$

$$y''=xe^{2x}, \quad y(0)=7, \quad y'(0)=1$$

$$y''=- x \sin x, \quad y(0)=1, \quad y'(0)=-3$$

$$y'''=x^2e^x, \quad y(0)=1, \quad y'(0)=-2, \quad y''(0)=3$$

$$y'''=2+\sin 2x, \quad y(0)=1, \quad y'(0)=-6, \quad y''(0)=3$$

$$y'''=2x+1, \quad y(2)=1, \quad y'(2)=-4, \quad y''(2)=7$$

[exer:1.2.5} Verify that the function is a solution of the initial value problem.

$$y=x\cos x; \quad y'=\cos x-y\tan x, \quad y(\pi/4)= {\pi\over4\sqrt{2}}$$

$${y={1+2\ln x\over x^2}+{1\over2}; \quad y'={x^2-2x^2y+2\over x^3}, \quad y(1)={3\over2}}$$

$$y= {\tan\left({x^2\over2}\right)}; \quad y'=x(1+y^2), \quad y(0)=0$$

$${y={2\over x-2}; \quad y'={-y(y+1)\over x}}, \quad y(1)=-2$$

[exer:1.2.6} Verify that the function is a solution of the initial value problem.

$$y=x^2(1+\ln x); \quad y''= {3xy'-4y\over x^2}, \quad y(e)=2e^2, \quad y'(e)=5e$$

$$y= {x^2\over3}+x-1; \quad y''= {x^2-xy'+y+1\over x^2}, \quad y(1)= {1\over3}, \quad y'(1)= {5\over3}$$

$$y=(1+x^2)^{-1/2}; \quad y''= {(x^2-1)y-x(x^2+1)y'\over (x^2+1)^2}, \quad y(0)=1,$$

$$y'(0)=0$$

$$y= {x^2\over 1-x}; \quad y''= {2(x+y)(xy'-y)\over x^3}, \quad y(1/2)=1/2, \quad y'(1/2)=3$$

[exer:1.2.7} Suppose an object is launched from a point 320 feet above the earth with an initial velocity of 128 ft/sec upward, and the only force acting on it thereafter is gravity. Take $$g=32$$ ft/sec$$^2$$.

Find the highest altitude attained by the object.

Determine how long it takes for the object to fall to the ground.

[exer:1.2.8} Let $$a$$ be a nonzero real number.

Verify that if $$c$$ is an arbitrary constant then $y=(x-c)^a \tag{A}$ is a solution of $y'=ay^{(a-1)/a} \tag{B}$ on $$(c,\infty)$$.

Suppose $$a<0$$ or $$a>1$$. Can you think of a solution of (B) that isn’t of the form (A)?

[exer:1.2.9} Verify that $y= \left\{ \begin{array}{cl} e^x-1,& x \ge 0, \\[6pt] 1-e^{-x},& x < 0, \end{array}\right.$ is a solution of $y'=|y|+1$ on $$(-\infty,\infty)$$.

[exer:1.2.10]

Verify that if $$c$$ is any real number then $y=c^2+cx+2c+1 \tag{A}$ satisfies $y'={-(x+2)+\sqrt{x^2+4x+4y}\over2} \tag{B}$ on some open interval. Identify the open interval.

Verify that $y_1={-x(x+4)\over4}$ also satisfies (B) on some open interval, and identify the open interval. (Note that $$y_1$$ can’t be obtained by selecting a value of $$c$$ in (A).)