(a) Transform into a single second-order equation

Given: [ x_1'=x_1-2x_2,\qquad x_2'=3x_1-4x_2 ]

From the first equation, [ 2x_2=x_1-x_1' ] so [ x_2=\frac{x_1-x_1'}{2} ]

Differentiate: [ x_2'=\frac{x_1'-x_1''}{2} ]

Substitute into the second equation: [ \frac{x_1'-x_1''}{2}=3x_1-4\left(\frac{x_1-x_1'}{2}\right) ] [ \frac{x_1'-x_1''}{2}=3x_1-2x_1+2x_1' ] [ \frac{x_1'-x_1''}{2}=x_1+2x_1' ]

Multiplying by (2): [ x_1'-x_1''=2x_1+4x_1' ] [ x_1''+3x_1'+2x_1=0 ]

Thus the required second-order equation is [ \boxed{x_1''+3x_1'+2x_1=0} ]

Initial conditions: [ x_1(0)=1 ] and [ x_1'(0)=x_1(0)-2x_2(0)=1-2(2)=-3 ]

So, [ \boxed{x_1(0)=1,\qquad x_1'(0)=-3} ]

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(b) Find (x_1) and (x_2)

The auxiliary equation is [ m^2+3m+2=0 ] [ (m+1)(m+2)=0 ] [ m=-1,\ -2 ]

Hence, [ x_1=c_1e^{-t}+c_2e^{-2t} ]

Using (x_1(0)=1): [ c_1+c_2=1 ]

Differentiate: [ x_1'=-c_1e^{-t}-2c_2e^{-2t} ]

Using (x_1'(0)=-3): [ -c_1-2c_2=-3 ] [ c_1+2c_2=3 ]

Solving: [ c_1+c_2=1 ] [ c_1+2c_2=3 ]

Subtracting gives [ c_2=2 ] and then [ c_1=-1 ]

Therefore, [ \boxed{x_1=-e^{-t}+2e^{-2t}} ]

Now, [ x_2=\frac{x_1-x_1'}{2} ]

Since [ x_1'=e^{-t}-4e^{-2t} ] we get [ x_2=\frac{(-e^{-t}+2e^{-2t})-(e^{-t}-4e^{-2t})}{2} ] [ x_2=\frac{-2e^{-t}+6e^{-2t}}{2} ] [ \boxed{x_2=-e^{-t}+3e^{-2t}} ]

Hence the required solution is [ \boxed{x_1(t)=-e^{-t}+2e^{-2t}} ] [ \boxed{x_2(t)=-e^{-t}+3e^{-2t}} ]

Check: [ x_1(0)=-1+2=1,\qquad x_2(0)=-1+3=2 ] which satisfy the given initial conditions.

First alternative

Given: [ y''+4y=2\sin 2t,\qquad y(0)=2,\qquad y'(0)=-1. ]

Homogeneous equation: [ y''+4y=0. ] Its solution is [ y_h=C_1\cos 2t+C_2\sin 2t. ]

Since the forcing term (2\sin 2t) is resonant, take [ y_p=t(A\cos 2t+B\sin 2t). ] Then [ y_p''+4y_p=-4A\sin 2t+4B\cos 2t. ] Comparing with (2\sin 2t), [ -4A=2,\qquad 4B=0. ] So [ A=-\frac12,\qquad B=0. ] Thus [ y_p=-\frac12 t\cos 2t. ]

General solution: [ y=C_1\cos 2t+C_2\sin 2t-\frac12 t\cos 2t. ]

Now apply the initial conditions.

[ y(0)=C_1=2. ]

Also, [ y'=-2C_1\sin 2t+2C_2\cos 2t-\frac12\cos 2t+t\sin 2t. ] At (t=0), [ y'(0)=2C_2-\frac12=-1. ] So [ 2C_2=-\frac12 \quad\Rightarrow\quad C_2=-\frac14. ]

Therefore the solution of the initial value problem is [ \boxed{y(t)=2\cos 2t-\frac14\sin 2t-\frac12 t\cos 2t.} ]

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OR alternative

Mass weighing (4) lb: [ m=\frac{W}{g}=\frac{4}{32}=\frac18\ \text{slug}. ] Spring stretch: [ s=1.5\text{ in}=\frac{1.5}{12}=\frac18\ \text{ft}. ] Spring constant: [ k=\frac{W}{s}=\frac{4}{1/8}=32\ \text{lb/ft}. ]

Let (u(t)) be displacement from equilibrium in feet. No damping, external force (2\cos 3t), so [ m u''+k u=2\cos 3t. ] Thus [ \frac18 u''+32u=2\cos 3t. ] Multiplying by (8), [ u''+256u=16\cos 3t. ]

Initial conditions: [ u(0)=\frac{2}{12}=\frac16\ \text{ft},\qquad u'(0)=0. ]

Homogeneous solution: [ u_h=C_1\cos 16t+C_2\sin 16t. ]

For the particular solution, try [ u_p=A\cos 3t+B\sin 3t. ] Substituting, [ (256-9)(A\cos 3t+B\sin 3t)=16\cos 3t. ] So [ 247A=16,\qquad B=0, ] giving [ A=\frac{16}{247}. ] Hence [ u_p=\frac{16}{247}\cos 3t. ]

General solution: [ u=C_1\cos 16t+C_2\sin 16t+\frac{16}{247}\cos 3t. ]

Using (u(0)=\frac16), [ C_1+\frac{16}{247}=\frac16 ] [ C_1=\frac16-\frac{16}{247}=\frac{151}{1482}. ]

Also, [ u'=-16C_1\sin 16t+16C_2\cos 16t-\frac{48}{247}\sin 3t. ] At (t=0), [ u'(0)=16C_2=0 \quad\Rightarrow\quad C_2=0. ]

Therefore, [ \boxed{u(t)=\frac{151}{1482}\cos 16t+\frac{16}{247}\cos 3t.} ]

Equivalently, [ \boxed{u(t)=\left(\frac16-\frac{16}{247}\right)\cos 16t+\frac{16}{247}\cos 3t.} ]

The differential equation is $y''+4y=t^2+3e^t.$

1. Homogeneous solution

Solve $y''+4y=0.$ Auxiliary equation: $r^2+4=0 \implies r=\pm 2i.$ So, $y_h=C_1\cos 2t+C_2\sin 2t.$

2. Particular solution

For (t^2), try $y_{p1}=At^2+Bt+C.$ Then $y_{p1}''=2A.$ Substitute: $2A+4(At^2+Bt+C)=t^2$ $4At^2+4Bt+(2A+4C)=t^2.$ Comparing coefficients: $4A=1 \implies A=\frac14,$ $4B=0 \implies B=0,$ $2A+4C=0 \implies \frac12+4C=0 \implies C=-\frac18.$ Thus, $y_{p1}=\frac14t^2-\frac18.$

For (3e^t), try $y_{p2}=De^t.$ Then $y_{p2}''+4y_{p2}=De^t+4De^t=5De^t.$ So, $5D=3 \implies D=\frac35.$ Thus, $y_{p2}=\frac35e^t.$

Hence the general solution is $y=C_1\cos 2t+C_2\sin 2t+\frac14t^2-\frac18+\frac35e^t.$

3. Apply initial conditions

Given (y(0)=0): $C_1-\frac18+\frac35=0$ $C_1=\frac18-\frac35=\frac{5-24}{40}=-\frac{19}{40}.$

Now differentiate: $y'=-2C_1\sin 2t+2C_2\cos 2t+\frac12t+\frac35e^t.$ Given (y'(0)=2): $2C_2+\frac35=2$ $2C_2=\frac75$ $C_2=\frac7{10}.$

Therefore, the solution of the initial value problem is $\boxed{y=-\frac{19}{40}\cos 2t+\frac7{10}\sin 2t+\frac14t^2-\frac18+\frac35e^t.}$

(a) Equation of the form $f\left(\dfrac{dy}{dx}, \dfrac{y}{x}\right)=0$

Write $p = \dfrac{dy}{dx}$. The equation relates $p$ and $y/x$ only, so it is homogeneous. Two cases:

Method 1 (solve for $p$). Solve $f(p, y/x)=0$ for $p$ to get $p = \phi\!\left(\dfrac{y}{x}\right)$. Put $y = vx$, so $\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$:

$v + x\frac{dv}{dx} = \phi(v)\;\Rightarrow\; \frac{dv}{\phi(v)-v} = \frac{dx}{x}$

The variables are separated. Integrate:

$\int\frac{dv}{\phi(v)-v} = \ln x + c$

then replace $v$ by $y/x$.

Method 2 (solve for $y/x$). If it is easier, solve for $\dfrac{y}{x} = \psi(p)$, i.e. $y = x\psi(p)$. Differentiating with respect to $x$:

$p = \psi(p) + x\psi'(p)\frac{dp}{dx}\;\Rightarrow\; \frac{dx}{x} = \frac{\psi'(p)\,dp}{p-\psi(p)}$

Integrating gives $x$ in terms of $p$ (with a constant $c$); together with $y = x\psi(p)$ this gives the solution in parametric form (eliminate $p$ if possible).

(b) $y = px + ap(1-p)$

This is Clairaut's equation $y = px + F(p)$ with $F(p) = ap(1-p)$.

Complete solution: replace $p$ by an arbitrary constant $c$:

$y = cx + ac(1-c)$

Singular solution: differentiate the complete solution with respect to $c$:

$0 = x + a(1-2c)\;\Rightarrow\; c = \frac{x+a}{2a}$

Substituting back, with $y = c(x+a) - ac^2$:

$y = \frac{(x+a)^2}{2a} - \frac{(x+a)^2}{4a} = \frac{(x+a)^2}{4a}$

$\boxed{(x+a)^2 = 4ay}$

This parabola is the envelope of the family of straight lines and is the singular solution.

Let $p = \partial z/\partial x$. The equation is

$4yz\,p = -(4+2y)\;\Rightarrow\; 2z\,\frac{\partial z}{\partial x} = -\frac{2+y}{y}\quad\Big(\text{i.e. } \frac{\partial (z^2)}{\partial x} = -\frac{2+y}{y}\Big)$

It contains no $q$, so $y$ acts as a constant and we integrate with respect to $x$:

$z^2 + \frac{2+y}{y}\,x = f(y)$

where $f$ is arbitrary (Lagrange's auxiliary equations $\dfrac{dx}{4yz} = \dfrac{dy}{0} = \dfrac{dz}{-(4+2y)}$ give $y=c_1$ and $z^2 + \frac{2+y}{y}x = c_2$). The integral surfaces are therefore $\phi\!\left(y,\ z^2 + \dfrac{2+y}{y}x\right)=0$.

Fitting the curve $y^2+z^2=1,\ x+z=2$.

On the curve, $x = 2 - z$ and $z^2 = 1-y^2$. Put $c = \dfrac{2+y}{y}$. Then

$c_2 = z^2 + c\,x = 1 - y^2 + c(2-z)\;\Rightarrow\; c_2 - (1-y^2) - 2c = -c\,z$

Squaring removes $z$ (using $z^2 = 1-y^2$ on the curve):

$\big[c_2 - (1-y^2) - 2c\big]^2 = c^2(1-y^2)$

Now replace $c_2$ by $z^2 + cx$ and $c$ by $\dfrac{2+y}{y}$:

$\big[z^2 + y^2 - 1 + c(x-2)\big]^2 = c^2(1-y^2)$

Multiplying by $y^2$:

$\boxed{\big[y(y^2+z^2-1) + (2+y)(x-2)\big]^2 = (2+y)^2(1-y^2)}$

Check: on the curve $y^2+z^2-1=0$ and $x-2=-z$, so the left side is $(2+y)^2z^2 = (2+y)^2(1-y^2)$, which equals the right side. The surface therefore contains the given circle and satisfies the differential equation (it is a function of $y$ and $z^2+\frac{2+y}{y}x$ only).

The given PDE is

$$z_{xx}+2z_{xy}+z_{yy}=0.$$

Here,

$$A=1,\quad B=2,\quad C=1.$$

So,

$$B^2-4AC=4-4=0.$$

Hence the equation is parabolic.

The characteristic equation is

$$A\left(\frac{dy}{dx}\right)^2-B\left(\frac{dy}{dx}\right)+C=0$$ $$\left(\frac{dy}{dx}\right)^2-2\left(\frac{dy}{dx}\right)+1=0$$ $$\left(\frac{dy}{dx}-1\right)^2=0.$$

So,

$$\frac{dy}{dx}=1$$

and the characteristic curves are

$$y-x=c \quad \text{or} \quad x-y=c.$$

Choose the new variables

$$\xi=x+y,\qquad \eta=x-y.$$

Then

$$\frac{\partial}{\partial x}=\frac{\partial}{\partial \xi}+\frac{\partial}{\partial \eta}, \qquad \frac{\partial}{\partial y}=\frac{\partial}{\partial \xi}-\frac{\partial}{\partial \eta}.$$

Therefore,

$$z_{xx}=z_{\xi\xi}+2z_{\xi\eta}+z_{\eta\eta},$$ $$z_{xy}=z_{\xi\xi}-z_{\eta\eta},$$ $$z_{yy}=z_{\xi\xi}-2z_{\xi\eta}+z_{\eta\eta}.$$

Substituting in the given equation,

$$z_{xx}+2z_{xy}+z_{yy} =4z_{\xi\xi}=0.$$

Thus the canonical form is

$$\boxed{z_{\xi\xi}=0}.$$

Now integrate with respect to (\xi):

$$z_{\xi}=\phi(\eta),$$

where (\phi) is an arbitrary function of (\eta). Again integrating,

$$z=\xi\phi(\eta)+\psi(\eta),$$

where (\psi) is another arbitrary function of (\eta).

Returning to the original variables (\xi=x+y,\ \eta=x-y),

$$\boxed{z=(x+y)\phi(x-y)+\psi(x-y)}.$$

Hence the general solution of the given PDE is

$$\boxed{z=(x+y)F(x-y)+G(x-y)}$$

where (F) and (G) are arbitrary functions.

Fourier Series of the Given Function

Given
[ f(x)= \begin{cases} x+1, & -1\le x<0,\ 1-x, & 0\le x<1, \end{cases} \qquad f(x+2)=f(x). ]

The function is periodic with period (2), so (2L=2\Rightarrow L=1).
Also, [ f(-x)=f(x), ] so (f(x)) is an even function. Hence all sine coefficients (b_n=0), and the Fourier series contains only cosine terms: [ f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}a_n\cos(n\pi x). ]

1. Find (a_0): [ a_0=\frac1L\int_{-L}^{L}f(x),dx =\int_{-1}^{1}f(x),dx. ] Since (f(x)) is even, [ a_0=2\int_0^1(1-x),dx =2\left[x-\frac{x^2}{2}\right]_0^1 =2\left(1-\frac12\right)=1. ] Thus, [ \frac{a_0}{2}=\frac12. ]

2. Find (a_n): [ a_n=\frac1L\int_{-1}^{1}f(x)\cos(n\pi x),dx =2\int_0^1(1-x)\cos(n\pi x),dx. ] Let [ I=\int_0^1(1-x)\cos(n\pi x),dx. ] Using integration by parts, [ I=\left[(1-x)\frac{\sin(n\pi x)}{n\pi}\right]_0^1 +\frac1{n\pi}\int_0^1\sin(n\pi x),dx. ] The boundary term is zero. Hence, [ I=\frac1{n\pi}\left[-\frac{\cos(n\pi x)}{n\pi}\right]_0^1 =\frac{1-\cos(n\pi)}{n^2\pi^2} =\frac{1-(-1)^n}{n^2\pi^2}. ] Therefore, [ a_n=2I=\frac{2(1-(-1)^n)}{n^2\pi^2}. ] For even (n), (a_n=0).
For odd (n), (a_n=\dfrac{4}{n^2\pi^2}).

3. Fourier series: [ \boxed{ f(x)=\frac12+\frac{4}{\pi^2}\sum_{k=0}^{\infty} \frac{\cos\big((2k+1)\pi x\big)}{(2k+1)^2} } ] or [ \boxed{ f(x)=\frac12+\frac{4}{\pi^2} \left( \frac{\cos\pi x}{1^2} +\frac{\cos3\pi x}{3^2} +\frac{\cos5\pi x}{5^2} +\cdots \right) } ]

Sketch of the graph for three periods

Over one period ([-1,1]), the function is [ f(x)=1-|x|. ] It has a peak of height (1) at (x=0) and zeros at (x=-1,1).
Because the period is (2), the graph repeats every (2) units. Thus:

• peaks occur at even integers: (x=\ldots,-2,0,2,\ldots)
• zeros occur at odd integers: (x=\ldots,-3,-1,1,3,\ldots)

For three periods, say from (x=-3) to (x=3), the graph is a triangular wave:

      /\        /\        /\
     /  \      /  \      /  \
____/    \____/    \____/    \____
-3   -2   -1    0    1    2    3

The straight-line segments join the points [ (-3,0),(-2,1),(-1,0),(0,1),(1,0),(2,1),(3,0). ]

Hence the required Fourier series is [ \boxed{ f(x)=\frac12+\frac{4}{\pi^2} \left( \cos\pi x+\frac{\cos3\pi x}{3^2} +\frac{\cos5\pi x}{5^2} +\cdots \right) } ] and the graph is the triangular wave shown above.

Let
[ u(x,y)=X(x)Y(y) ] be a separated solution of Laplace’s equation [ u_{xx}+u_{yy}=0. ] Substituting gives [ X''Y+XY''=0 ] or [ \frac{X''}{X}=-\frac{Y''}{Y}=\mu^2 \quad \text{(separation constant)}. ] Hence [ X''-\mu^2 X=0,\qquad Y''+\mu^2 Y=0. ]

The boundary conditions in the (y)-direction are [ u(x,0)=u(x,b)=0, ] so [ Y(0)=Y(b)=0. ] The non-trivial solutions of [ Y''+\mu^2Y=0 ] with these boundary conditions are [ Y_n(y)=\sin\left(\frac{n\pi y}{b}\right),\qquad \mu=\frac{n\pi}{b},\quad n=1,2,3,\dots ]

For each (n), the equation for (X) becomes [ X_n''-\left(\frac{n\pi}{b}\right)^2X_n=0. ] Its general solution is [ X_n(x)=A_n e^{n\pi x/b}+B_n e^{-n\pi x/b}. ] Using the boundary condition [ u(0,y)=0 \implies X_n(0)=0, ] we get (A_n+B_n=0), so [ X_n(x)=C_n\sinh\left(\frac{n\pi x}{b}\right). ]

Therefore the separated solutions are [ u_n(x,y)=C_n\sinh\left(\frac{n\pi x}{b}\right) \sin\left(\frac{n\pi y}{b}\right). ] By superposition, [ u(x,y)=\sum_{n=1}^{\infty}C_n\sinh\left(\frac{n\pi x}{b}\right) \sin\left(\frac{n\pi y}{b}\right). ]

Now apply the remaining boundary condition [ u(a,y)=f(y). ] Then [ f(y)=\sum_{n=1}^{\infty}C_n\sinh\left(\frac{n\pi a}{b}\right) \sin\left(\frac{n\pi y}{b}\right). ] This is the Fourier sine series of (f(y)) on (0<y<b). Hence [ C_n\sinh\left(\frac{n\pi a}{b}\right)

\frac{2}{b}\int_0^b f(y)\sin\left(\frac{n\pi y}{b}\right),dy. ] Thus [ \boxed{ C_n= \frac{2}{b\sinh\left(\frac{n\pi a}{b}\right)} \int_0^b f(y)\sin\left(\frac{n\pi y}{b}\right),dy } ] and the required solution is [ \boxed{ u(x,y)=\sum_{n=1}^{\infty}C_n \sinh\left(\frac{n\pi x}{b}\right) \sin\left(\frac{n\pi y}{b}\right) } ] with (C_n) as above. This solution satisfies all the given Dirichlet boundary conditions, provided (f) is piecewise smooth and (f(0)=f(b)=0).

First alternative

Given $y''+4y=2\sin 2t,\qquad y(0)=2,\quad y'(0)=-1.$

Homogeneous equation: $y''+4y=0.$ Its characteristic equation is $r^2+4=0 \implies r=\pm 2i.$ So $y_h=C_1\cos 2t+C_2\sin 2t.$

Since the forcing term (2\sin 2t) is already in (y_h), take $y_p=t(A\cos 2t+B\sin 2t).$

Differentiate: $y_p'=A\cos 2t+B\sin 2t+t(-2A\sin 2t+2B\cos 2t),$ $y_p''=-4A\sin 2t+4B\cos 2t-4t(A\cos 2t+B\sin 2t).$

Substitute into (y''+4y=2\sin 2t): $-4A\sin 2t+4B\cos 2t=2\sin 2t.$ Hence $-4A=2 \implies A=-\frac12,\qquad 4B=0 \implies B=0.$ Thus $y_p=-\frac{t}{2}\cos 2t.$

General solution: $y(t)=C_1\cos 2t+C_2\sin 2t-\frac{t}{2}\cos 2t.$

Use (y(0)=2): $C_1=2.$

Now $y'(t)=-2C_1\sin 2t+2C_2\cos 2t-\frac12\cos 2t+t\sin 2t.$ Using (y'(0)=-1): $2C_2-\frac12=-1 \implies 2C_2=-\frac12 \implies C_2=-\frac14.$

Therefore, $\boxed{y(t)=2\cos 2t-\frac14\sin 2t-\frac{t}{2}\cos 2t.}$

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Second alternative

Spring constant: $k=\frac{F}{x}=\frac{2}{0.10}=20\ \text{N/m}.$

Mass: $m=2\ \text{kg}.$

Damping coefficient: $c=\frac{F}{v}=\frac{4}{1}=4\ \text{N s/m}.$

External force: $F(t)=4\cos 2t.$

Equation of motion: $m x''+c x'+k x=F(t),$ so $2x''+4x'+20x=4\cos 2t.$ Dividing by (2): $x''+2x'+10x=2\cos 2t.$

Homogeneous equation: $x''+2x'+10x=0.$ Characteristic equation: $r^2+2r+10=0 \implies r=-1\pm 3i.$ Hence $x_h=e^{-t}(C_1\cos 3t+C_2\sin 3t).$

For the particular solution, take $x_p=A\cos 2t+B\sin 2t.$ Then $x_p'=-2A\sin 2t+2B\cos 2t,$ $x_p''=-4A\cos 2t-4B\sin 2t.$

Substitute into (x''+2x'+10x=2\cos 2t): $(6A+4B)\cos 2t+(-4A+6B)\sin 2t=2\cos 2t.$ Thus $6A+4B=2,\qquad -4A+6B=0.$ Solving, $A=\frac{3}{13},\qquad B=\frac{2}{13}.$ So $x_p=\frac{3}{13}\cos 2t+\frac{2}{13}\sin 2t.$

General solution: $x(t)=e^{-t}(C_1\cos 3t+C_2\sin 3t)+\frac{3}{13}\cos 2t+\frac{2}{13}\sin 2t.$

Initial conditions: initially at rest at equilibrium means $x(0)=0,\qquad x'(0)=0.$

From (x(0)=0): $C_1+\frac{3}{13}=0 \implies C_1=-\frac{3}{13}.$

Now $x'(t)=e^{-t}\left[(-C_1+3C_2)\cos 3t+(-C_2-3C_1)\sin 3t\right]-\frac{6}{13}\sin 2t+\frac{4}{13}\cos 2t.$ Using (x'(0)=0): $-C_1+3C_2+\frac{4}{13}=0.$ Since (C_1=-\frac{3}{13}), $\frac{3}{13}+3C_2+\frac{4}{13}=0 \implies 3C_2=-\frac{7}{13} \implies C_2=-\frac{7}{39}.$

Therefore, $\boxed{x(t)=e^{-t}\left(-\frac{3}{13}\cos 3t-\frac{7}{39}\sin 3t\right)+\frac{3}{13}\cos 2t+\frac{2}{13}\sin 2t.}$

The position is in metres.

Transient part: $x_{\text{tr}}(t)=e^{-t}\left(-\frac{3}{13}\cos 3t-\frac{7}{39}\sin 3t\right).$

Steady-state part: $x_{\text{ss}}(t)=\frac{3}{13}\cos 2t+\frac{2}{13}\sin 2t.$

Equivalently, $x_{\text{ss}}(t)=\frac{\sqrt{13}}{13}\cos(2t-\delta),$ where $\delta=\tan^{-1}\left(\frac{2}{3}\right).$

The differential equation is $y''+y'-2y=2t,\qquad y(0)=0,\quad y'(0)=1.$

1. Complementary function

Solve the homogeneous equation: $y''+y'-2y=0.$

Auxiliary equation: $m^2+m-2=0$ $(m+2)(m-1)=0$ $m=1,\,-2.$

Hence, $y_c=C_1e^t+C_2e^{-2t}.$

2. Particular integral

Since the RHS is (2t), take $y_p=At+B.$

Then $y_p'=A,\qquad y_p''=0.$

Substitute into the differential equation: $0+A-2(At+B)=2t$ $-2At+(A-2B)=2t.$

Comparing coefficients: $-2A=2 \implies A=-1,$ $A-2B=0 \implies -1-2B=0 \implies B=-\frac12.$

Thus, $y_p=-t-\frac12.$

3. General solution

$y=y_c+y_p$ $y=C_1e^t+C_2e^{-2t}-t-\frac12.$

4. Apply initial conditions

Given (y(0)=0): $C_1+C_2-\frac12=0$ $C_1+C_2=\frac12. \quad (1)$

Now $y'=C_1e^t-2C_2e^{-2t}-1.$

Given (y'(0)=1): $C_1-2C_2-1=1$ $C_1-2C_2=2. \quad (2)$

Subtracting (2) from (1): $(C_1+C_2)-(C_1-2C_2)=\frac12-2$ $3C_2=-\frac32$ $C_2=-\frac12.$

From (1): $C_1-\frac12=\frac12 \implies C_1=1.$

Therefore, $\boxed{y=e^t-\frac12e^{-2t}-t-\frac12}$

This is the required solution of the given initial value problem.

(a) Method for solving Clairaut’s equation

The Clairaut equation is
[ y=px+f(p),\qquad p=\frac{dy}{dx}. ]

Step 1: Differentiate both sides with respect to (x):
[ p=p+x\frac{dp}{dx}+f'(p)\frac{dp}{dx}. ]

Step 2: Cancel (p) from both sides:
[ \left[x+f'(p)\right]\frac{dp}{dx}=0. ]

So either
[ \frac{dp}{dx}=0 ] or
[ x+f'(p)=0. ]

Step 3: If (\frac{dp}{dx}=0), then (p=c), where (c) is an arbitrary constant. Substituting (p=c) in the original equation gives the complete solution:
[ y=cx+f(c). ]

Step 4: If (x+f'(p)=0), eliminate (p) between this equation and the original equation. The resulting relation, free from arbitrary constants, is the singular solution. It is the envelope of the family (y=cx+f(c)).

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(b) Complete and singular solution of

[ y=px+ap(1-p) ]

This is a Clairaut equation with
[ f(p)=ap(1-p). ]

Complete solution

Replace (p) by the arbitrary constant (c):
[ y=cx+ac(1-c). ]

Hence the complete solution is
[ \boxed{y=cx+ac-ac^2} ] or
[ \boxed{y=cx+ac(1-c)}. ]

Singular solution

For the singular solution, use
[ x+f'(p)=0. ]

Here
[ f(p)=ap(1-p)=ap-ap^2, ] so
[ f'(p)=a-2ap=a(1-2p). ]

Thus
[ x+a(1-2p)=0 ] [ x+a-2ap=0 ] [ 2ap=x+a ] [ p=\frac{x+a}{2a}. ]

Substitute this value of (p) into the original equation:
[ y=px+ap(1-p). ]

So
[ y=x\left(\frac{x+a}{2a}\right)+a\left(\frac{x+a}{2a}\right)\left(1-\frac{x+a}{2a}\right). ]

Now
[ 1-\frac{x+a}{2a}=\frac{2a-x-a}{2a}=\frac{a-x}{2a}. ]

Therefore
[ y=\frac{x(x+a)}{2a}+\frac{(x+a)(a-x)}{4a}. ]

Taking L.C.M. (4a),
[ y=\frac{2x(x+a)+(x+a)(a-x)}{4a}. ]

Simplify:
[ 2x(x+a)+(x+a)(a-x) =(x+a)\left[2x+a-x\right] =(x+a)(x+a) =(x+a)^2. ]

Hence
[ y=\frac{(x+a)^2}{4a}. ]

Therefore the singular solution is
[ \boxed{y=\frac{(x+a)^2}{4a}}. ]

So, for (a\neq 0),
[ \boxed{\text{Complete solution: } y=cx+ac(1-c)} ] [ \boxed{\text{Singular solution: } y=\frac{(x+a)^2}{4a}}. ]

If (a=0), the equation reduces to (y=px), whose complete solution is (y=cx), and there is no separate singular solution.

Setting up. The given one-parameter family is

$z=cxy(x^2+y^2)\quad\Longrightarrow\quad u(x,y,z)=\frac{z}{xy(x^2+y^2)}=c.$

Its normal direction is $(u_x,u_y,u_z)$. With $D=xy(x^2+y^2)=x^3y+xy^3$:

$u_x=-\frac{z\,y(3x^2+y^2)}{D^2},\qquad u_y=-\frac{z\,x(x^2+3y^2)}{D^2},\qquad u_z=\frac1D,$

so the normal is proportional to

$P=-zy(3x^2+y^2),\qquad Q=-zx(x^2+3y^2),\qquad R=xy(x^2+y^2).$

A surface $z=\phi(x,y)$ cuts every member of the family orthogonally when its normal $(p,q,-1)$ is perpendicular to $(P,Q,R)$, i.e. when it satisfies the Lagrange equation

$Pp+Qq=R.$

Auxiliary equations.

$\frac{dx}{-zy(3x^2+y^2)}=\frac{dy}{-zx(x^2+3y^2)}=\frac{dz}{xy(x^2+y^2)}.$

First integral. Taking the first two ratios:

$x(x^2+3y^2)\,dx=y(3x^2+y^2)\,dy .$

Put $u=x+y,\ v=x-y$. Then $x^3+3xy^2=\tfrac12(u^3+v^3)$ and $3x^2y+y^3=\tfrac12(u^3-v^3)$, and the equation becomes

$u^3\,dv+v^3\,du=0\ \Rightarrow\ \frac{du}{u^3}+\frac{dv}{v^3}=0\ \Rightarrow\ \frac1{u^2}+\frac1{v^2}=\text{const}.$

Since $u^2+v^2=2(x^2+y^2)$ and $uv=x^2-y^2$,

$\frac{x^2+y^2}{(x^2-y^2)^2}=C_1 .$

Second integral. Use multipliers $x,\ y,\ 4z$. Then

$x\,P+y\,Q=-4zxy(x^2+y^2)=-4z\,R,$

so

$\frac{x\,dx+y\,dy}{-4z\,xy(x^2+y^2)}=\frac{dz}{xy(x^2+y^2)}\ \Rightarrow\ x\,dx+y\,dy+4z\,dz=0,$

$x^2+y^2+4z^2=C_2 .$

General orthogonal surface.

$x^2+y^2+4z^2=\varphi\!\left[\frac{x^2+y^2}{(x^2-y^2)^2}\right].$

Condition. The surface must pass through $x^2-y^2=a^2,\ z=0$. On this curve $C_2=x^2+y^2$ and $C_1=\dfrac{x^2+y^2}{a^4}$, so $C_2=a^4C_1$, i.e. $\varphi(t)=a^4t$.

Required surface:

$\boxed{(x^2+y^2+4z^2)(x^2-y^2)^2=a^4(x^2+y^2)}$

Let $F(x,y,z,p,q)=p^2x+q^2y-z=0$, where $p=\partial z/\partial x,\ q=\partial z/\partial y$.

Charpit's auxiliary equations

$\frac{dx}{F_p}=\frac{dy}{F_q}=\frac{dz}{pF_p+qF_q}=\frac{-dp}{F_x+pF_z}=\frac{-dq}{F_y+qF_z}.$

Here $F_p=2px,\ F_q=2qy,\ F_x=p^2,\ F_y=q^2,\ F_z=-1$, so $pF_p+qF_q=2(p^2x+q^2y)=2z$, $F_x+pF_z=p^2-p$ and $F_y+qF_z=q^2-q$:

$\frac{dx}{2px}=\frac{dy}{2qy}=\frac{dz}{2z}=\frac{-dp}{p(p-1)}=\frac{-dq}{q(q-1)}.$

Finding $p$ and $q$. From $\dfrac{dx}{2px}=\dfrac{-dp}{p(p-1)}$:

$\frac{dx}{2x}=-\frac{dp}{p-1}\ \Rightarrow\ \tfrac12\ln x=-\ln(p-1)+\ln a\ \Rightarrow\ p=1+\frac{a}{\sqrt x}.$

In the same way, from $\dfrac{dy}{2qy}=\dfrac{-dq}{q(q-1)}$:

$q=1+\frac{b}{\sqrt y}.$

Check against the PDE. With these values, $p^2x=(\sqrt x+a)^2$ and $q^2y=(\sqrt y+b)^2$, so $p^2x+q^2y=z$ gives $z=(\sqrt x+a)^2+(\sqrt y+b)^2$. Here $a$ and $b$ are independent constants, so this is a two-parameter solution. It can be confirmed by integrating $dz=p\,dx+q\,dy$.

Integrating $dz=p\,dx+q\,dy$

$dz=\left(1+\frac{a}{\sqrt x}\right)dx+\left(1+\frac{b}{\sqrt y}\right)dy$

$z=x+y+2a\sqrt x+2b\sqrt y+C.$

Substituting $p,q$ in $p^2x+q^2y=z$:

$z=(x+2a\sqrt x+a^2)+(y+2b\sqrt y+b^2)=x+y+2a\sqrt x+2b\sqrt y+a^2+b^2,$

so $C=a^2+b^2$.

Complete integral:

$\boxed{z=x+y+2a\sqrt x+2b\sqrt y+a^2+b^2=(\sqrt x+a)^2+(\sqrt y+b)^2}$

where $a,b$ are arbitrary constants.

Sketch of $f(x)$ over three periods (period $2L$, shown on $-3L\le x\le 3L$)

On $-L\le x\le 0$ the graph is the straight line $y=x+L$, rising from $0$ at $x=-L$ to $L$ at $x=0$. On $0<x<L$ it is the horizontal line $y=L$. This pattern repeats every $2L$:

 f
 L |      ____       ____       ____
   |     /    |     /    |     /    |
   |    /     |    /     |    /     |
 0 |___/      |___/      |___/      |___
   -3L  -2L  -L   0    L   2L   3L

Each period is a ramp from $0$ to $L$ followed by a flat level at $L$. The function is continuous at $x=0,\pm2L,\dots$ and has a jump from $L$ down to $0$ at $x=\pm L,\pm3L,\dots$

Fourier series. Period $2L$:

$f(x)=\frac{a_0}{2}+\sum_{n=1}^{\infty}\left(a_n\cos\frac{n\pi x}{L}+b_n\sin\frac{n\pi x}{L}\right).$

Constant term.

$a_0=\frac1L\left[\int_{-L}^{0}(x+L)\,dx+\int_0^L L\,dx\right]=\frac1L\left[\frac{L^2}{2}+L^2\right]=\frac{3L}{2},\qquad \frac{a_0}{2}=\frac{3L}{4}.$

Cosine coefficients. Write $k=n\pi/L$. The integral over $(0,L)$ of $L\cos kx$ is $0$. Integrating by parts,

$$a_n=\frac1L\int_{-L}^{0}(x+L)\cos kx\,dx=\frac1L\left[\frac{(x+L)\sin kx}{k}\right]_{-L}^{0}-\frac1{Lk}\int_{-L}^{0}\sin kx\,dx =\frac{1-(-1)^n}{L\,k^2}$$

$a_n=\frac{L\,[1-(-1)^n]}{n^2\pi^2}\quad\left(=\frac{2L}{n^2\pi^2}\text{ for odd }n,\ 0\text{ for even }n\right).$

Sine coefficients.

$b_n=\frac1L\left[\int_{-L}^{0}(x+L)\sin kx\,dx+\int_0^L L\sin kx\,dx\right].$

First integral: $\left[-\dfrac{(x+L)\cos kx}{k}\right]_{-L}^{0}+\dfrac1k\int_{-L}^0\cos kx\,dx=-\dfrac Lk$. Second integral: $\dfrac{L[1-(-1)^n]}{k}$. Hence

$b_n=\frac1L\cdot\frac{L}{k}\left(-1+1-(-1)^n\right)=\frac{(-1)^{n+1}L}{n\pi}.$

Result

$\boxed{f(x)=\frac{3L}{4}+\sum_{n=1}^{\infty}\left[\frac{L\,(1-(-1)^n)}{n^2\pi^2}\cos\frac{n\pi x}{L}+\frac{(-1)^{n+1}L}{n\pi}\sin\frac{n\pi x}{L}\right]}$

Check at $x=0$ (continuity point): $\frac{3L}{4}+\frac{2L}{\pi^2}\sum_{n\ \text{odd}}\frac1{n^2}=\frac{3L}{4}+\frac{2L}{\pi^2}\cdot\frac{\pi^2}{8}=L=f(0)$. At the jump $x=\pm L$ the series converges to $L/2$, the mean of $L$ and $0$.

---

OR: eigenvalues and eigenfunctions of $y''+\lambda y=0,\ y'(0)=0,\ y(\pi)=0$

Case $\lambda=0$: $y=A+Bx$. $y'(0)=B=0$ and $y(\pi)=A=0$, so only $y\equiv0$. Not an eigenvalue.

Case $\lambda=-\mu^2<0$: $y=A\cosh\mu x+B\sinh\mu x$. $y'(0)=B\mu=0\Rightarrow B=0$, and $y(\pi)=A\cosh\mu\pi=0\Rightarrow A=0$. No eigenvalue.

Case $\lambda=\mu^2>0$: $y=A\cos\mu x+B\sin\mu x$, $y'=-A\mu\sin\mu x+B\mu\cos\mu x$.

$y'(0)=B\mu=0\Rightarrow B=0$. Then $y(\pi)=A\cos\mu\pi=0$ with $A\ne0$ requires $\cos\mu\pi=0$:

$\mu\pi=\frac{(2n-1)\pi}{2}\ \Rightarrow\ \mu=\frac{2n-1}{2},\quad n=1,2,3,\dots$

$\boxed{\lambda_n=\left(\frac{2n-1}{2}\right)^2=\frac{(2n-1)^2}{4},\qquad y_n(x)=\cos\frac{(2n-1)x}{2}}\quad(n=1,2,\dots)$

The eigenvalues are $\tfrac14,\tfrac94,\tfrac{25}{4},\dots$, and each eigenfunction is determined up to a constant multiple.