(a) Reduction to 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 this and (x_2=\frac{x_1-x_1'}{2}) into the second equation: [ \frac{x_1'-x_1''}{2}=3x_1-4\left(\frac{x_1-x_1'}{2}\right). ]

Simplify the right-hand side: [ 3x_1-2(x_1-x_1')=3x_1-2x_1+2x_1'=x_1+2x_1'. ]

Thus [ \frac{x_1'-x_1''}{2}=x_1+2x_1'. ]

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

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

Also, from (x_2=\frac{x_1-x_1'}{2}), once (x_1) is known, (x_2) follows.

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(b) Solution satisfying the initial conditions

The initial conditions are [ x_1(0)=1,\qquad x_2(0)=2. ]

From (x_1'=x_1-2x_2), [ x_1'(0)=x_1(0)-2x_2(0)=1-4=-3. ]

So we solve [ x_1''+3x_1'+2x_1=0,\qquad x_1(0)=1,\quad x_1'(0)=-3. ]

The characteristic equation is [ r^2+3r+2=0 ] [ (r+1)(r+2)=0. ]

Thus [ r=-1,,-2. ]

Therefore, [ x_1(t)=C_1e^{-t}+C_2e^{-2t}. ]

Using (x_1(0)=1): [ C_1+C_2=1. \quad (1) ]

Differentiate: [ x_1'(t)=-C_1e^{-t}-2C_2e^{-2t}. ]

Using (x_1'(0)=-3): [ -C_1-2C_2=-3 ] or [ C_1+2C_2=3. \quad (2) ]

Subtracting (1) from (2): [ C_2=2. ]

Then from (1): [ C_1=-1. ]

Hence [ \boxed{x_1(t)=-e^{-t}+2e^{-2t}}. ]

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

We have [ x_1'=e^{-t}-4e^{-2t}. ]

Therefore [ x_2(t)=\frac{(-e^{-t}+2e^{-2t})-(e^{-t}-4e^{-2t})}{2} ] [ =\frac{-2e^{-t}+6e^{-2t}}{2} ] [ =-e^{-t}+3e^{-2t}. ]

So [ \boxed{x_2(t)=-e^{-t}+3e^{-2t}}. ]

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Final answer: [ \boxed{x_1(t)=-e^{-t}+2e^{-2t}} ] [ \boxed{x_2(t)=-e^{-t}+3e^{-2t}} ]

These satisfy (x_1(0)=1) and (x_2(0)=2).

The maximum loan is the present value of an ordinary annuity of 1000 per month for 30 years.

Given: monthly payment $R = 1000$; monthly rate $i = \dfrac{0.10}{12} = 0.0083333$; number of payments $n = 30 \times 12 = 360$.

$PV = R\left[\frac{1-(1+i)^{-n}}{i}\right] = 1000\left[\frac{1-(1.0083333)^{-360}}{0.0083333}\right]$

Since $(1.0083333)^{360} \approx 19.837$, we have $(1.0083333)^{-360} \approx 0.05041$. Then

$PV = 1000\times\frac{1-0.05041}{0.0083333} = 1000 \times 113.95 \approx 113{,}951$

Maximum mortgage loan $\approx$ 113,951 (in the same currency as the 1000 monthly payment, which the question gives in dollars).

Taking the first alternative:

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

Homogeneous equation: [ y''+4y=0. ] Characteristic equation: [ r^2+4=0 \implies r=\pm 2i. ] So [ y_c=c_1\cos 2t+c_2\sin 2t. ]

Since the right-hand side is (\sin 2t), which is already a solution of the homogeneous equation, we try [ y_p=t(A\cos 2t+B\sin 2t). ] Differentiating, [ y_p''+4y_p=-4A\sin 2t+4B\cos 2t. ] Comparing with (2\sin 2t), [ -4A=2 \implies A=-\frac12,\qquad 4B=0 \implies B=0. ] Hence [ y_p=-\frac{t}{2}\cos 2t. ]

Therefore the general solution is [ y=c_1\cos 2t+c_2\sin 2t-\frac{t}{2}\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. ] Thus [ 2c_2=-\frac12 \implies c_2=-\frac14. ]

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

Check: [ y(0)=2,\qquad y'(0)=-\frac12-\frac12=-1. ] Hence the required solution is verified.

For the differential equation [ y''+4y=t^2+3e^t, ] first solve the homogeneous equation: [ y''+4y=0. ] Its characteristic equation is [ r^2+4=0 \implies r=\pm 2i. ] So, [ y_h(t)=C_1\cos 2t+C_2\sin 2t. ]

Now find a particular solution.

For (t^2), take [ y_{p1}=At^2+Bt+C. ] Then [ y_{p1}''=2A. ] Substitute into (y''+4y=t^2): [ 2A+4(At^2+Bt+C)=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}=\frac{t^2}{4}-\frac18. ]

For (3e^t), take [ y_{p2}=De^t. ] Then [ y_{p2}''+4y_{p2}=De^t+4De^t=5De^t. ] So, [ 5D=3 \implies D=\frac35. ] Hence, [ y_{p2}=\frac35e^t. ]

Therefore the general solution is [ y(t)=C_1\cos 2t+C_2\sin 2t+\frac{t^2}{4}-\frac18+\frac35e^t. ]

Now apply the initial conditions.

[ y(0)=0 ] gives [ C_1-\frac18+\frac35=0. ] So, [ C_1=\frac18-\frac35=-\frac{19}{40}. ]

Also, [ y'(t)=-2C_1\sin 2t+2C_2\cos 2t+\frac{t}{2}+\frac35e^t. ] Using (y'(0)=2): [ 2C_2+\frac35=2, ] so [ 2C_2=\frac75, ] [ C_2=\frac7{10}. ]

Hence the solution of the initial value problem is [ \boxed{ y(t)=-\frac{19}{40}\cos 2t+\frac7{10}\sin 2t+\frac{t^2}{4}-\frac18+\frac35e^t } ]

(a) Solving a homogeneous differential equation of the form

[ f\left(\frac{dy}{dx},\frac{y}{x}\right)=0 ]

A first-order homogeneous differential equation can be written as
[ \frac{dy}{dx}=F\left(\frac{y}{x}\right) ] or, more generally, as a relation between (\frac{dy}{dx}) and (\frac{y}{x}).

Method:

1. Put
   [ y=vx ] where (v) is a new variable.

2. Differentiate with respect to (x): [ \frac{dy}{dx}=v+x\frac{dv}{dx}. ]

3. Substitute (y=vx) and (\frac{dy}{dx}=v+x\frac{dv}{dx}) into the given equation. It becomes an equation involving (v) and (x) only: [ f\left(v+x\frac{dv}{dx},,v\right)=0. ]

4. Solve this equation for (\frac{dv}{dx}). Usually it takes the separable form
   [ \frac{dv}{dx}=\phi(v). ]

5. Separate the variables: [ \frac{dv}{\phi(v)}=\frac{dx}{x}. ]

6. Integrate both sides: [ \int \frac{dv}{\phi(v)}=\int \frac{dx}{x}+C. ]

7. Finally, replace (v) by (\frac{y}{x}) to get the required general solution.

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(b) Complete and singular solution of

[ y=px+ap(1-p) ] where
[ p=\frac{dy}{dx}. ]

This is a Clairaut’s equation of the form
[ y=px+f(p), ] with
[ f(p)=ap(1-p)=ap-ap^2. ]

Complete solution

For Clairaut’s equation, the complete solution is obtained by replacing (p) by an arbitrary constant (c). Hence
[ \boxed{y=cx+ac(1-c)} ] or
[ \boxed{y=cx+ac-ac^2}. ]

Singular solution

Differentiate the given equation with respect to (p):
[ \frac{dy}{dp}=x+f'(p). ] But for the singular solution, we use
[ x+f'(p)=0. ]

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

Thus,
[ x+a(1-2p)=0 ] [ x=-a(1-2p) ] [ x=a(2p-1). ]

Hence,
[ 2p-1=\frac{x}{a} ] [ p=\frac{x+a}{2a}. ]

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

We have
[ p=\frac{x+a}{2a}. ]

Then
[ px=\frac{x(x+a)}{2a} ] and
[ 1-p=1-\frac{x+a}{2a}=\frac{a-x}{2a}. ]

Therefore,
[ ap(1-p)=a\cdot \frac{x+a}{2a}\cdot \frac{a-x}{2a} =\frac{(x+a)(a-x)}{4a} =\frac{a^2-x^2}{4a}. ]

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

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

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

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Final answers:

[ \boxed{\text{Complete solution: } y=cx+ac(1-c)} ]

[ \boxed{\text{Singular solution: } y=\frac{(x+a)^2}{4a}} ]

Write the equation as $F(x,y,z,p,q) = 4yzp + 4 + 2y = 0$, where $p = \partial z/\partial x$ and $q = \partial z/\partial y$. It contains no $q$, so $F_q = 0$.

Characteristic equations (Lagrange–Charpit)

$\frac{dx}{F_p} = \frac{dy}{F_q} = \frac{dz}{pF_p + qF_q} = \frac{dx}{4yz} = \frac{dy}{0} = \frac{dz}{4yzp}$

From $dy/0$ we get $dy = 0$, so $y$ = constant along each characteristic. Then $dz = p\,dx$, and the equation itself gives

$zp = -\frac{4+2y}{4y} = -\frac{2+y}{2y}.$

Integration

With $y$ held constant, $zp = \tfrac12\,\dfrac{\partial (z^2)}{\partial x}$, so

$\frac{\partial (z^2)}{\partial x} = -\frac{2+y}{y} \quad\Longrightarrow\quad z^2 = -\frac{(2+y)}{y}\,x + h(y)$

where $h(y)$ is arbitrary (a complete integral in terms of the characteristic parameter). This is the general solution.

Applying the initial curve

The curve is $y^2 + z^2 = 1$, $x + z = 2$. Put $z = w = \sqrt{1-y^2}$ (taking the branch of the circle used) and $x = 2 - w$ on it:

$1 - y^2 = -\frac{(2+y)(2-w)}{y} + h(y) \;\Longrightarrow\; h(y) = 1 - y^2 + \frac{(2+y)\,(2-w)}{y}.$

Substituting back:

$z^2 = 1 - y^2 + \frac{(2+y)\left(2 - x - \sqrt{1-y^2}\right)}{y}$

or

$\boxed{\,y z^2 = y(1-y^2) + (2+y)\left(2 - x - \sqrt{1-y^2}\right)\,}$

Check: on the curve $x = 2 - z$ with $z = \sqrt{1-y^2}$, the bracket $2 - x - \sqrt{1-y^2} = z - z = 0$, so $yz^2 = y(1-y^2)$, i.e. $z^2 = 1 - y^2$, as required. This is the equation of the required integral surface.

The given PDE is

$$\frac{\partial^2 z}{\partial x^2}+2\frac{\partial^2 z}{\partial x\partial y}+\frac{\partial^2 z}{\partial y^2}=0.$$

Comparing with

$$A z_{xx}+2B z_{xy}+C z_{yy}=0,$$

we get

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

Hence

$$B^2-AC=1-1=0,$$

so the equation is parabolic.

The characteristic equation is

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

i.e.

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

Thus

$$\left(\frac{dy}{dx}-1\right)^2=0 \quad\Rightarrow\quad \frac{dy}{dx}=1.$$

So the characteristic curves are

$$y-x=c.$$

Let

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

Then

$$z_x=-z_\xi+z_\eta,\qquad z_y=z_\xi.$$

Also,

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

Substituting in the given PDE,

$$(z_{\xi\xi}-2z_{\xi\eta}+z_{\eta\eta}) +2(-z_{\xi\xi}+z_{\xi\eta})+z_{\xi\xi}=0.$$

The terms containing (z_{\xi\xi}) and (z_{\xi\eta}) cancel, leaving

$$z_{\eta\eta}=0.$$

This is the required canonical form.

Now integrate (z_{\eta\eta}=0) with respect to (\eta):

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

where (\phi) is an arbitrary function of (\xi). Again integrating with respect to (\eta),

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

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

Reverting to (x,y), since (\xi=y-x) and (\eta=x),

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

or equivalently,

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

Hence the general solution is

$$\boxed{z=x f(y-x)+g(y-x)}$$

where (f) and (g) are arbitrary functions.

Sketch

The function is $f(x) = 1 - |x|$ on $[-1,1]$, repeated with period 2. Its graph is a zig-zag (triangular wave):

• peaks of height 1 at $x = 0, \pm 2, \pm 4, \dots$,
• zeros at $x = \pm1, \pm3, \dots$,
• straight lines between them (slope $+1$ on $[-1,0]$ and $-1$ on $[0,1]$).

Over three periods, from $x=-3$ to $x=3$, the graph rises from 0 at $x=-3$ to a peak of 1 at $x=-2$, falls to 0 at $x=-1$, rises to 1 at $x=0$, falls to 0 at $x=1$, rises to 1 at $x=2$ and falls to 0 at $x=3$.

Fourier series

Period $2L = 2$, so $L = 1$. The function is even ($f(-x) = f(x)$), hence $b_n = 0$.

$a_0 = \frac{2}{L}\int_0^1 (1-x)\,dx = 2\left[x - \frac{x^2}{2}\right]_0^1 = 1$

$a_n = 2\int_0^1 (1-x)\cos n\pi x\,dx = 2\left[\frac{(1-x)\sin n\pi x}{n\pi}\right]_0^1 + \frac{2}{n\pi}\int_0^1 \sin n\pi x\,dx = \frac{2\,(1-\cos n\pi)}{n^2\pi^2}$

So $a_n = \dfrac{4}{n^2\pi^2}$ for odd $n$ and $a_n = 0$ for even $n$.

$f(x) = \frac{a_0}{2} + \sum a_n\cos n\pi x = \frac12 + \frac{4}{\pi^2}\left[\cos\pi x + \frac{\cos 3\pi x}{3^2} + \frac{\cos 5\pi x}{5^2} + \cdots\right]$

Check at $x=0$: $\tfrac12 + \tfrac{4}{\pi^2}\cdot\tfrac{\pi^2}{8} = 1 = f(0)$.

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OR: Solve $r = a^2 t$ by Monge's method

Here $R = 1,\ S = 0,\ T = -a^2$ (with $r = z_{xx}$, $t = z_{yy}$). Monge's equations are

$R\,dy^2 - S\,dx\,dy + T\,dx^2 = 0 \quad\text{and}\quad R\,dp\,dy + T\,dq\,dx = 0.$

The first gives $dy^2 - a^2dx^2 = 0$, i.e. $dy = \pm a\,dx$, so $y \mp ax =$ constant.

• For $dy = a\,dx$: the second equation becomes $a\,dp - a^2\,dq = 0$, so $dp = a\,dq$ and $p - aq = F(y - ax)$.
• For $dy = -a\,dx$: $-a\,dp - a^2\,dq = 0$, so $p + aq = G(y + ax)$.

Solving for $p$ and $q$ gives $p = \tfrac12(F+G)$ and $q = \tfrac1{2a}(G-F)$. Integrating $dz = p\,dx + q\,dy$:

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

where $\phi$ and $\psi$ are arbitrary functions. Check: $z_{xx} = a^2(\phi'' + \psi'')$ and $z_{yy} = \phi'' + \psi''$, so $r = a^2 t$.

Main part: Two-dimensional Laplace equation in a rectangle

We have to solve

$$u_{xx}+u_{yy}=0,\qquad 0<x<a,\quad 0<y<b$$

with boundary conditions

$$u(x,0)=0,\qquad u(x,b)=0,\qquad 0<x<a,$$ $$u(0,y)=0,\qquad u(a,y)=f(y),\qquad 0\le y\le b.$$

Use the method of separation of variables. Let

$$u(x,y)=X(x)Y(y).$$

Substituting into Laplace’s equation,

$$X''Y+XY''=0$$

so that

$$\frac{X''}{X}=-\frac{Y''}{Y}=-\lambda.$$

Thus,

$$X''+\lambda X=0,\qquad Y''-\lambda Y=0.$$

The boundary conditions in the (y)-direction are homogeneous:

$$Y(0)=0,\qquad Y(b)=0.$$

For nontrivial solutions we must take (\lambda<0). Put

$$\lambda=-k^2.$$

Then

$$Y''+k^2Y=0.$$

Its general solution is

$$Y(y)=A\cos ky+B\sin ky.$$

Using (Y(0)=0),

$$A=0.$$

Hence

$$Y(y)=B\sin ky.$$

Using (Y(b)=0),

$$B\sin kb=0.$$

For a nontrivial solution, (B\ne0), so

$$\sin kb=0.$$

Therefore,

$$kb=n\pi,\qquad n=1,2,3,\dots$$

so

$$k=\frac{n\pi}{b}.$$

Thus the (y)-eigenfunctions are

$$Y_n(y)=\sin\frac{n\pi y}{b}.$$

Now the (x)-equation becomes

$$X''-\frac{n^2\pi^2}{b^2}X=0.$$

Its general solution is

$$X(x)=C_1\cosh\frac{n\pi x}{b}+C_2\sinh\frac{n\pi x}{b}.$$

Using the boundary condition (u(0,y)=0), we get (X(0)=0). Hence

$$C_1=0.$$

Therefore,

$$X_n(x)=C_2\sinh\frac{n\pi x}{b}.$$

So the elementary solutions are

$$u_n(x,y)=A_n\sinh\frac{n\pi x}{b}\sin\frac{n\pi y}{b}.$$

By the principle of superposition, the required solution is

$$u(x,y)=\sum_{n=1}^{\infty}A_n\sinh\frac{n\pi x}{b}\sin\frac{n\pi y}{b}.$$

Now apply the non-homogeneous boundary condition

$$u(a,y)=f(y).$$

Then

$$f(y)=\sum_{n=1}^{\infty}A_n\sinh\frac{n\pi a}{b}\sin\frac{n\pi y}{b}.$$

This is the Fourier sine series of (f(y)) on (0<y<b). Hence

$$A_n\sinh\frac{n\pi a}{b} = \frac{2}{b}\int_0^b f(y)\sin\frac{n\pi y}{b}\,dy.$$

Therefore,

$$A_n= \frac{2}{b\sinh\frac{n\pi a}{b}} \int_0^b f(y)\sin\frac{n\pi y}{b}\,dy.$$

Hence the solution of the given boundary value problem is

$$\boxed{ u(x,y)= \sum_{n=1}^{\infty} \left[ \frac{2}{b\sinh\frac{n\pi a}{b}} \int_0^b f(\eta)\sin\frac{n\pi \eta}{b}\,d\eta \right] \sinh\frac{n\pi x}{b} \sin\frac{n\pi y}{b} }$$

where (\eta) is a dummy variable. This satisfies Laplace’s equation and all the given boundary conditions. For a classical solution, (f) should be piecewise smooth and satisfy the compatibility conditions (f(0)=f(b)=0).

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OR part: Non-dimensionalization of the wave equation

Let

$$S=\frac{x}{L}.$$

Then

$$x=LS.$$

Write

$$u(x,t)=U(S,t).$$

By the chain rule,

$$u_x=\frac{\partial U}{\partial S}\frac{dS}{dx} =\frac{1}{L}U_S.$$

Differentiating again,

$$u_{xx}=\frac{1}{L^2}U_{SS}.$$

The wave equation is

$$u_{tt}=a^2u_{xx}.$$

Substituting (u_{xx}=\dfrac{1}{L^2}U_{SS}),

$$U_{tt}=\frac{a^2}{L^2}U_{SS}.$$

Multiplying by (L^2),

$$L^2U_{tt}=a^2U_{SS}.$$

Rename (U) again as (u). Then

$$\boxed{L^2u_{tt}=a^2u_{SS}}$$

or, writing (S=s),

$$\boxed{L^2u_{tt}=a^2u_{ss}}.$$

Thus the wave equation becomes non-dimensional with respect to the spatial variable (S=x/L).
If the printed statement writes (L^2u_{ss}=a^2u_{tt}), then the subscripts (s) and (t) are interchanged; the correct transformed form is (L^2u_{tt}=a^2u_{ss}).

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

Homogeneous equation: $y''+4y=0$ Auxiliary equation: $r^2+4=0 \implies r=\pm 2i$ So, $y_h=c_1\cos 2t+c_2\sin 2t.$

Particular solution: Since the forcing term is (2\sin t), try $y_p=A\sin t+B\cos t.$ Then $y_p''=-A\sin t-B\cos t.$ Substitute into (y''+4y=2\sin t): $(-A\sin t-B\cos t)+4(A\sin t+B\cos t)=2\sin t$ $3A\sin t+3B\cos t=2\sin t.$ Comparing coefficients: $3A=2 \implies A=\frac23,\qquad 3B=0 \implies B=0.$ Hence, $y_p=\frac23\sin t.$

General solution: $y(t)=c_1\cos 2t+c_2\sin 2t+\frac23\sin t.$

Now apply the initial conditions.

$y(0)=c_1=2.$

Also, $y'(t)=-2c_1\sin 2t+2c_2\cos 2t+\frac23\cos t.$ Using (y'(0)=-1): $2c_2+\frac23=-1$ $2c_2=-\frac53$ $c_2=-\frac56.$

Therefore the solution of the initial value problem is $\boxed{y(t)=2\cos 2t-\frac56\sin 2t+\frac23\sin t}.$

The differential equation is

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

Homogeneous solution

Auxiliary equation:

$m^2+m-2=0$

$(m+2)(m-1)=0$

$m=1,\ -2.$

So,

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

Particular solution

Since the right-hand side 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:

$-2A=2 \implies A=-1,$

$A-2B=0 \implies -1-2B=0 \implies B=-\frac12.$

Hence,

$y_p=-t-\frac12.$

Therefore the general solution is

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

Using initial conditions

$y(0)=C_1+C_2-\frac12=0$

$C_1+C_2=\frac12. \quad (1)$

Also,

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

$y'(0)=C_1-2C_2-1=1$

$C_1-2C_2=2. \quad (2)$

Subtracting (1) from (2):

$3C_2=-\frac32 \implies C_2=-\frac12.$

Then from (1):

$C_1-\frac12=\frac12 \implies C_1=1.$

Final answer

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

For Clairaut's equation
$y = px + f(p), \qquad p=\frac{dy}{dx},$ differentiate both sides with respect to $x$: $\frac{dy}{dx}=p+x\frac{dp}{dx}+f'(p)\frac{dp}{dx}.$ Since $\frac{dy}{dx}=p$, this gives $p=p+\bigl[x+f'(p)\bigr]\frac{dp}{dx}$ so that $\bigl[x+f'(p)\bigr]\frac{dp}{dx}=0.$

Hence two cases arise:

1. General solution:
   $\frac{dp}{dx}=0 \implies p=c \quad (\text{arbitrary constant}).$ Substituting $p=c$ in the original equation, $y=cx+f(c).$ This is the required general solution.

2. Singular solution:
   $x+f'(p)=0.$ Eliminate $p$ between this equation and the original equation $y=px+f(p)$. The resulting relation between $x$ and $y$, free from the arbitrary constant $c$, is the singular solution.

Thus the complete solution consists of the general solution $y=cx+f(c)$ and, if it exists, the singular solution obtained by eliminating $p$ from $x+f'(p)=0,\qquad y=px+f(p).$

Let
[ p=\frac{dy}{dx}. ] The given equation is [ y=px+ap(1-p). ] This is a Clairaut-type equation of the form [ y=px+f(p),\qquad f(p)=ap(1-p). ]

Complete solution

For Clairaut’s equation, replace (p) by an arbitrary constant (c). Hence the complete solution is [ \boxed{y=cx+ac(1-c)}. ]

Singular solution

Write [ F(x,y,p)=y-px-ap(1-p)=0. ] For the singular solution, eliminate (p) using [ \frac{\partial F}{\partial p}=0. ] Now, [ \frac{\partial F}{\partial p}=-x-a(1-2p)=0. ] So [ x=a(2p-1) ] or [ p=\frac{x+a}{2a}. ] Substitute this value of (p) into the complete solution: [ y=cx+ac(1-c), ] where (c=p=\dfrac{x+a}{2a}). Thus [ y=\frac{x+a}{2a}x+a\frac{x+a}{2a}\left(1-\frac{x+a}{2a}\right). ] Simplifying, [ y=\frac{(x+a)^2}{4a}. ]

Therefore the singular solution is [ \boxed{y=\frac{(x+a)^2}{4a}},\qquad a\ne 0. ]

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

Step 1: Differential equation of the given system. The one-parameter family is

$z = c\,xy(x^2+y^2) \;\Rightarrow\; c = \frac{z}{xy(x^2+y^2)}$

Differentiating $z = c(x^3y + xy^3)$:

$p_1 = c(3x^2y + y^3) = \frac{z(3x^2+y^2)}{x(x^2+y^2)},\qquad q_1 = c(x^3 + 3xy^2) = \frac{z(x^2+3y^2)}{y(x^2+y^2)}$

The normal to the family is $(p_1, q_1, -1)$.

Step 2: Orthogonal surfaces. A surface $z=f(x,y)$ is orthogonal to the family if its normal $(p,q,-1)$ is perpendicular to $(p_1,q_1,-1)$:

$p_1 p + q_1 q + 1 = 0\ \Rightarrow\ p_1 p + q_1 q = -1$

This is a Lagrange equation, with auxiliary equations

$\frac{dx}{p_1} = \frac{dy}{q_1} = \frac{dz}{-1}$

Step 3: First integral. From $\dfrac{dx}{p_1} = \dfrac{dy}{q_1}$:

$\frac{x\,dx}{3x^2+y^2} = \frac{y\,dy}{x^2+3y^2}\ \Rightarrow\ (x^3+3xy^2)\,dx = (3x^2y+y^3)\,dy$

Put $u=x^2,\ v=y^2$: $(u+3v)\,du = (3u+v)\,dv$. With $s=u+v$ and $d=u-v$, this is $2s\,dd = d\,ds$, giving $d^2 = c_1 s$:

$\frac{(x^2-y^2)^2}{x^2+y^2} = c_1$

Step 4: Second integral. Note that $x p_1 + y q_1 = \dfrac{z[(3x^2+y^2)+(x^2+3y^2)]}{x^2+y^2} = 4z$. Using multipliers $x, y$:

$\frac{x\,dx + y\,dy}{4z} = \frac{dz}{-1}\ \Rightarrow\ x\,dx+y\,dy+4z\,dz = 0\ \Rightarrow\ x^2+y^2+4z^2 = c_2$

Step 5: Apply the condition. The surface must pass through $x^2-y^2=a^2,\ z=0$. On this curve, with $x^2+y^2=s$:

$c_1 = \frac{a^4}{s},\qquad c_2 = s \quad\Rightarrow\quad c_1c_2 = a^4$

Hence the required orthogonal surface is

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

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:

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

Here $F_x = p^2,\ F_y = q^2,\ F_z = -1,\ F_p = 2px,\ F_q = 2qy$, so

$\frac{dp}{p^2-p} = \frac{dq}{q^2-q} = \frac{dz}{-2(p^2x+q^2y)} = \frac{dz}{-2z} = \frac{dx}{-2px} = \frac{dy}{-2qy}$

Finding an integral. Take $\dfrac{dp}{p(p-1)} = \dfrac{dx}{-2px}$, which gives $\dfrac{dp}{p-1} = -\dfrac{dx}{2x}$, and take $\dfrac{dp}{p(p-1)} = \dfrac{dz}{-2z}$. Using these two,

$\frac{2\,dp}{p} + \frac{dx}{x} - \frac{dz}{z} = 0\ \Rightarrow\ p^2x = a^2 z \qquad (a\ \text{constant})$

Verification: along the auxiliary equations $dp:dx:dz = (p^2-p) : (-2px) : (-2z)$, so $2(p-1) - 2p + 2 = 0$, as required.

Finding $p$ and $q$.

$p = a\sqrt{\frac{z}{x}}$

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

$q = \sqrt{1-a^2}\,\sqrt{\frac{z}{y}}$

Solving $dz = p\,dx + q\,dy$:

$\frac{dz}{\sqrt z} = a\,\frac{dx}{\sqrt x} + \sqrt{1-a^2}\,\frac{dy}{\sqrt y}$

Integrating,

$2\sqrt z = 2a\sqrt x + 2\sqrt{1-a^2}\sqrt y + 2b$

The complete integral is therefore

$\boxed{\sqrt z = a\sqrt x + \sqrt{1-a^2}\,\sqrt y + b}\quad\text{or}\quad z = \big(a\sqrt x + \sqrt{1-a^2}\sqrt y + b\big)^2$

where $a$ and $b$ are arbitrary constants.

Fourier series of $f(x) = \begin{cases} x+L, & -L\le x\le 0 \\ L, & 0<x<L\end{cases}$, $\ f(x+2L)=f(x)$.

Graph (three periods). On $[-L,0]$ the graph is a straight line of slope 1 rising from $0$ at $x=-L$ to $L$ at $x=0$. On $(0,L)$ it is the horizontal line $y=L$. At $x=L$ the function drops from $L$ back to $0$ (a jump discontinuity), and the same pattern repeats on every interval of length $2L$, for example on $[-3L,-L]$, $[-L,L]$ and $[L,3L]$. The curve is continuous at $x=0$, $\pm2L$, ... and has jumps at $x=\pm L,\pm3L,\dots$

Coefficients.

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

$a_n = \frac1L\left[\int_{-L}^{0}(x+L)\cos\frac{n\pi x}{L}\,dx + \int_0^L L\cos\frac{n\pi x}{L}\,dx\right]$

The second integral is zero. In the first, put $u=x+L$, so $\cos\frac{n\pi x}{L} = (-1)^n\cos\frac{n\pi u}{L}$:

$a_n = \frac{(-1)^n}{L}\int_0^L u\cos\frac{n\pi u}{L}\,du = \frac{(-1)^n}{L}\cdot\frac{L^2}{n^2\pi^2}\big[(-1)^n-1\big] = \frac{L\,[1-(-1)^n]}{n^2\pi^2}$

so $a_n = \dfrac{2L}{n^2\pi^2}$ for odd $n$ and $0$ for even $n$.

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

$\int_{-L}^{0}(x+L)\sin\frac{n\pi x}{L}\,dx = (-1)^n\int_0^L u\sin\frac{n\pi u}{L}\,du = (-1)^n\cdot\left(-\frac{L^2(-1)^n}{n\pi}\right) = -\frac{L^2}{n\pi}$

$L\int_0^L\sin\frac{n\pi x}{L}\,dx = \frac{L^2}{n\pi}\big[1-(-1)^n\big]$

$b_n = \frac1L\left[-\frac{L^2}{n\pi}+\frac{L^2}{n\pi}-\frac{L^2(-1)^n}{n\pi}\right] = \frac{(-1)^{n+1}L}{n\pi}$

Fourier series:

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

Check at $x=0$: $\dfrac{3L}{4}+\dfrac{2L}{\pi^2}\left(1+\dfrac1{9}+\dfrac1{25}+\cdots\right) = \dfrac{3L}{4}+\dfrac{2L}{\pi^2}\cdot\dfrac{\pi^2}{8} = L = f(0)$.

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

• $\lambda = 0$: $y = Ax+B$. $y'(0)=A=0$ and $y(\pi)=B=0$, so only the trivial solution.
• $\lambda<0$ (let $\lambda=-k^2$): $y = A\cosh kx + B\sinh kx$. $y'(0)=Bk=0\Rightarrow B=0$, and $y(\pi)=A\cosh k\pi=0\Rightarrow A=0$: trivial.
• $\lambda>0$ (let $\lambda=k^2$, $k>0$): $y = A\cos kx + B\sin kx$. $y'(0)=Bk=0\Rightarrow B=0$. Then $y(\pi)=A\cos k\pi=0$ with $A\ne0$ requires $k\pi = \dfrac{(2n-1)\pi}{2}$.

Eigenvalues:

$\lambda_n = \left(\frac{2n-1}{2}\right)^2,\quad n=1,2,3,\dots$

Eigenfunctions:

$y_n(x) = \cos\frac{(2n-1)x}{2},\quad n=1,2,3,\dots$