Mathematics II (BT-202) - Unit 3 Short Notes

How unit 3 is examined

Formation of PDEs by eliminating constants or arbitrary functions, Lagrange and Charpit solution methods, and homogeneous linear PDEs with constant coefficients; the last two topics carry most of the marks.

Formulation of Partial Differential equations

<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">Medium weight</span>

Definition. A partial differential equation (PDE) is an equation containing an unknown function of two or more variables and its partial derivatives. Notation: $p=\dfrac{\partial z}{\partial x}$, $q=\dfrac{\partial z}{\partial y}$.

Key points.

1. To form a PDE, eliminate the arbitrary constants or the arbitrary function from the given relation by differentiating partially.
2. Eliminating two arbitrary constants gives a first-order PDE (or a second-order one if more differentiation is needed).
3. Eliminating one arbitrary function $\phi$ from $\phi(u,v)=0$ gives a first-order PDE, and the order equals the number of functions eliminated.
4. For $z=f(u)$ with one function, differentiate w.r.t. $x$ and $y$ and eliminate $f'$ by taking the ratio of $p$ and $q$.
5. For $f(u,v)=0$, where $u,v$ contain $z$, use $u_x=\partial u/\partial x+p\,\partial u/\partial z$ and similarly for $y$, and eliminate through the determinant $u_xv_y-u_yv_x=0$.

Example. $z=f(y/x)$: $p=f'\left(-\frac{y}{x^2}\right)$, $q=f'\cdot\frac1x$. Dividing, $p/q=-y/x$.

==Eliminating an arbitrary function gives a first-order PDE: $z=f(y/x)$ gives $xp+yq=0$.==

Example 2. $z=y^2+2f\left(\frac1x+\log y\right)$: $p=-\frac{2f'}{x^2}$, $q=2y+\frac{2f'}{y}$. Since $2f'=-x^2p$, $q=2y-\frac{x^2p}{y}$, so $x^2p+yq=2y^2$.

Example 3. $f(u,v)=0$ with $u=x^2+y^2+z^2$, $v=z^2-2xy$: $u_x=2x+2zp$, $u_y=2y+2zq$, $v_x=-2y+2zp$, $v_y=-2x+2zq$. Then $u_xv_y-u_yv_x=0$ gives $(x+zp)(zq-x)-(y+zq)(zp-y)=0$, i.e. $(x+y)\,z(q-p)=x^2-y^2$, so $z(p-q)=y-x$.

Answer frame. Open with "Let $z=f(u)$; differentiate partially w.r.t. $x$ and $y$"; write $p$ and $q$, then eliminate the arbitrary function by ratio or determinant; close with the boxed first-order PDE.

Asked: [7 marks] (Jun 2022, Jun 2025) Form the PDE by eliminating the arbitrary function from $Z=f(y/x)$; also from $z=y^2+2f\left(\frac1x+\log y\right)$. Asked: [7 marks] (Dec 2024) Construct a PDE from the relation $f(x^2+y^2+z^2,\,z^2-2xy)=0$.

Linear and Non-Linear Partial Differential Equations

<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">High weight</span>

Definition. A first-order PDE $f(x,y,z,p,q)=0$ is linear if $p,q$ occur only to the first degree, as in Lagrange's form $Pp+Qq=R$ ($P,Q,R$ functions of $x,y,z$); otherwise it is non-linear.

Key points.

1. Lagrange's equation $Pp+Qq=R$ is solved through the auxiliary equations $\dfrac{dx}{P}=\dfrac{dy}{Q}=\dfrac{dz}{R}$.
2. Two independent integrals $u=c_1$ and $v=c_2$ are needed, found by grouping variables or by multipliers $l,m,n$ with $\dfrac{l\,dx+m\,dy+n\,dz}{lP+mQ+nR}$ where $lP+mQ+nR=0$.
3. The general solution is $f(u,v)=0$, which may also be written $u=\phi(v)$.
4. For a non-linear equation, a complete integral has two arbitrary constants; Charpit's method finds it.
5. Charpit: write $f(x,y,z,p,q)=0$ and solve $\dfrac{dp}{f_x+pf_z}=\dfrac{dq}{f_y+qf_z}=\dfrac{dz}{-pf_p-qf_q}=\dfrac{dx}{-f_p}=\dfrac{dy}{-f_q}$.
6. Pick the easiest pair to get one relation between $p$ and $q$ (containing a constant $a$), solve for $p,q$, and integrate $dz=p\,dx+q\,dy$.
7. Equations like $x^2p^2+y^2q^2=z^2$ are made standard by $X=\ln x$, $Y=\ln y$, $Z=\ln z$, giving $P^2+Q^2=1$ with solution $Z=aX+bY+c$.

==Charpit's method: solve the auxiliary equations for one relation between $p$ and $q$, then integrate $dz=p\,dx+q\,dy$.==

Example 1. $(x-y)p+(x+y)q=2xz$. Auxiliary: $\frac{dx}{x-y}=\frac{dy}{x+y}=\frac{dz}{2xz}$. Multipliers $(1,1,0)$: $\frac{dx+dy}{2x}=\frac{dz}{2xz}$, so $\frac{dz}{z}=dx+dy$ and $ze^{-(x+y)}=c_2$. Multipliers $(x,y,0)$ and $(-y,x,0)$ both give denominator $x^2+y^2$: $\frac{x\,dx+y\,dy}{x^2+y^2}=\frac{x\,dy-y\,dx}{x^2+y^2}$, so $\tfrac12\ln(x^2+y^2)-\tan^{-1}\frac yx=c_1$. Solution: $f\left(\tfrac12\ln(x^2+y^2)-\tan^{-1}\frac yx,\; ze^{-(x+y)}\right)=0$.

Example 2. $(y+z)p+(x+z)q=x+y$. Using $(1,-1,0)$ and $(0,1,-1)$: $\frac{dx-dy}{-(x-y)}=\frac{dy-dz}{-(y-z)}$, so $\frac{x-y}{y-z}=c_1$. Using $(1,1,1)$: $\frac{d(x+y+z)}{2(x+y+z)}=\frac{d(x-y)}{-(x-y)}$, so $(x-y)^2(x+y+z)=c_2$. Solution: $f\left(\frac{x-y}{y-z},(x-y)^2(x+y+z)\right)=0$.

Example 3 (Charpit). $(p^2+q^2)y=qz$: $f_x=0$, $f_y=p^2+q^2$, $f_z=-q$, $f_p=2py$, $f_q=2qy-z$. Then $\frac{dp}{-pq}=\frac{dq}{p^2}$ gives $p\,dp+q\,dq=0$, so $p^2+q^2=a^2$. The PDE gives $q=\frac{a^2y}{z}$, $p=\frac{a}{z}\sqrt{z^2-a^2y^2}$. Then $z\,dz-a^2y\,dy=a\sqrt{z^2-a^2y^2}\,dx$, i.e. $\frac{d(z^2-a^2y^2)}{2\sqrt{z^2-a^2y^2}}=a\,dx$. Complete integral: $z^2-a^2y^2=(ax+b)^2$.

Example 4 (Charpit). $px+qy=pq$: $f_x=p$, $f_y=q$, $f_z=0$, $f_p=x-q$, $f_q=y-p$. Then $\frac{dp}{p}=\frac{dq}{q}$ gives $p=aq$. The PDE gives $q=\frac{ax+y}{a}$, $p=ax+y$, so $dz=\frac{(ax+y)\,d(ax+y)}{a}$. Complete integral: $z=\frac{(ax+y)^2}{2a}+b$.

Example 5. $x^2p^2+y^2q^2=z^2$ becomes $P^2+Q^2=1$ with $P=\frac xz p$, $Q=\frac yz q$. Put $Z=aX+bY+c$, $a^2+b^2=1$. Solution: $\ln z=a\ln x+\sqrt{1-a^2}\,\ln y+c$.

Answer frame. Lagrange: open with "Here $P=\dots$, $Q=\dots$, $R=\dots$; auxiliary equations are ..."; find $c_1$, then $c_2$, each with its multipliers stated; close with $f(u,v)=0$. Charpit: open with $f=0$ and its five derivatives; write the Charpit equations, derive the $p$-$q$ relation, solve for $p,q$, integrate $dz$; close with the complete integral.

Pitfall: In Charpit, taking the ratio $\frac{dp}{f_x+pf_z}=\frac{dq}{f_y+qf_z}$ and cancelling a wrong term is the usual slip, so write all five derivatives first.

Asked: [7 marks] (Dec 2023, Jun 2025) Solve $(x-y)p+(x+y)q=2xz$; also $(y+z)p+(x+z)q=x+y$. Asked: [7 marks] (Dec 2023, Jun 2023) Solve $(p^2+q^2)y=qz$ by Charpit's method. Asked: [7 marks] (Nov 2022) Solve $x^2p^2+y^2q^2=z^2$. Asked: [7 marks] (Dec 2024) Solve by Charpit's method $px+qy=pq$.

Homogeneous Linear Partial Differential Equations with Constants Coefficients

<span style="display:inline-block;padding:.16em .6em;border:1.5px solid currentColor;border-radius:999px;font-size:.68em;font-weight:700;letter-spacing:.06em;text-transform:uppercase;opacity:.75">High weight</span>

Definition. With $D=\partial/\partial x$ and $D'=\partial/\partial y$, a homogeneous linear PDE with constant coefficients is $F(D,D')z=f(x,y)$ where every term of $F$ has the same total degree. Its solution is $z=\text{C.F.}+\text{P.I.}$

Key points.

1. For the complementary function put $D=m$, $D'=1$ in $F$ and solve the auxiliary equation $F(m,1)=0$.
2. For distinct roots $m_1,m_2,\dots$: $\text{C.F.}=\phi_1(y+m_1x)+\phi_2(y+m_2x)+\dots$
3. For a repeated root $m$ (twice): $\text{C.F.}=\phi_1(y+mx)+x\phi_2(y+mx)$.
4. For $f=e^{ax+by}$: $\text{P.I.}=\dfrac{e^{ax+by}}{F(a,b)}$, provided $F(a,b)\neq0$.
5. For $f=\sin(ax+by)$ or $\cos(ax+by)$: replace $D^2\to-a^2$, $DD'\to-ab$, $D'^2\to-b^2$, and divide if the result is nonzero.
6. For a polynomial $f$: expand $F^{-1}$ as a series in $D'/D$ (or $D/D'$) and integrate term by term.
7. If the denominator becomes zero (resonance), factor $F$ and use $\dfrac{1}{(D-mD')^r}\,g(ax+by)=\dfrac{x^r}{r!}\,g$ for the vanishing factor, with the other factors evaluated at $(a,b)$.

==Solution $z=\text{C.F.}+\text{P.I.}$ where C.F. comes from $F(m,1)=0$ with roots giving $\phi(y+mx)$.==

Example 1. $(D^3-3D^2D'+4D'^3)z=e^{x+2y}$. Auxiliary: $m^3-3m^2+4=(m+1)(m-2)^2=0$, so $m=-1,2,2$. $\text{P.I.}=\dfrac{e^{x+2y}}{1-6+32}=\dfrac{e^{x+2y}}{27}$. $z=\phi_1(y-x)+\phi_2(y+2x)+x\phi_3(y+2x)+\tfrac1{27}e^{x+2y}$.

Example 2. $(D^2-4DD'+4D'^2)z=\cos(x-2y)$. $m=2,2$, so C.F. $=\phi_1(y+2x)+x\phi_2(y+2x)$. With $a=1,b=-2$: $D^2=-1$, $DD'=2$, $D'^2=-4$, so $F=-1-8-16=-25$. $z=\phi_1(y+2x)+x\phi_2(y+2x)-\tfrac1{25}\cos(x-2y)$.

Example 3. $(D^2+4DD'-5D'^2)z=\sin(2x+3y)$. $m=1,-5$; $F=-4-24+45=17$. $z=\phi_1(y+x)+\phi_2(y-5x)+\tfrac1{17}\sin(2x+3y)$.

Example 4. $(D^2-6DD'+9D'^2)z=12x^2+36xy$. $m=3,3$. $\text{P.I.}=\frac1{D^2}\left(1-\frac{3D'}D\right)^{-2}(12x^2+36xy)=\frac1{D^2}\left[12x^2+36xy+6\cdot18x^2\right]=\frac1{D^2}(120x^2+36xy)=10x^4+6x^3y$. $z=\phi_1(y+3x)+x\phi_2(y+3x)+10x^4+6x^3y$.

Example 5. $z_{xx}-z_{yy}=x^2y$: $(D^2-D'^2)z=x^2y$, $m=\pm1$. $\text{P.I.}=\frac1{D^2}\left(1-\frac{D'^2}{D^2}\right)^{-1}x^2y=\frac1{D^2}x^2y=\frac{x^4y}{12}$ since $D'^2(x^2y)=0$. $z=\phi_1(y+x)+\phi_2(y-x)+\frac{x^4y}{12}$.

Example 6. $(D^3-4D^2D'+4DD'^2)z=\cos(2x+y)$: $m(m-2)^2=0$, $m=0,2,2$, so C.F. $=\phi_1(y)+\phi_2(y+2x)+x\phi_3(y+2x)$. With $a=2,b=1$, $F=-8D+16D'$ gives $8(2D'-D)$, which is $0$ at $(2,1)$: resonance. So $\text{P.I.}=\dfrac{1}{D(D-2D')^2}\cos(2x+y)$; the factor $D\to 2i$ and $(D-2D')^2$ gives $\frac{x^2}{2}$, so $\text{P.I.}=\dfrac{x^2}{4}\sin(2x+y)$. $z=\phi_1(y)+\phi_2(y+2x)+x\phi_3(y+2x)+\tfrac{x^2}{4}\sin(2x+y)$.

Answer frame. Open with "Write the equation as $F(D,D')z=f$; auxiliary equation is $F(m,1)=0$"; state the roots and C.F.; then find P.I. with the rule for the type of $f$, showing the substituted value of $F$; close with $z=\text{C.F.}+\text{P.I.}$

Pitfall: For sine and cosine the substitution is $D^2\to-a^2$, $DD'\to-ab$, $D'^2\to-b^2$, not $D\to a$; using $D\to a$ gives the wrong sign and value.

Asked: [7 marks] (Jun 2022) Solve $(D^3-3D^2D'+4D'^3)z=e^{x+2y}$. Asked: [7 marks] (Nov 2022, Dec 2023, Dec 2024) Solve $(D^2-4DD'+4D'^2)Z=\cos(x-2y)$; also $(D^2+4DD'-5D'^2)Z=\sin(2x+3y)$; also $(D^3-4D^2D'+4DD'^2)Z=\cos(2x+y)$. Asked: [7 marks] (Jun 2023, Jun 2025) Solve $(D^2-6DD'+9D'^2)z=12x^2+36xy$; also $\frac{\partial^2z}{\partial x^2}-\frac{\partial^2z}{\partial y^2}=x^2y$.

Last-minute revision

• PDE notation: $p=z_x$, $q=z_y$; one arbitrary function gives a first-order PDE.
• $z=f(y/x)$ gives $xp+yq=0$; $f(x^2+y^2+z^2,z^2-2xy)=0$ gives $z(p-q)=y-x$.
• Lagrange: $Pp+Qq=R$, auxiliary $\frac{dx}P=\frac{dy}Q=\frac{dz}R$, solution $f(u,v)=0$.
• Charpit: $\frac{dp}{f_x+pf_z}=\frac{dq}{f_y+qf_z}=\frac{dz}{-pf_p-qf_q}=\frac{dx}{-f_p}=\frac{dy}{-f_q}$; then $dz=p\,dx+q\,dy$.
• $(p^2+q^2)y=qz$: $z^2-a^2y^2=(ax+b)^2$.
• $px+qy=pq$: $z=\frac{(ax+y)^2}{2a}+b$.
• $x^2p^2+y^2q^2=z^2$: $\ln z=a\ln x+\sqrt{1-a^2}\ln y+c$.
• C.F.: root $m$ gives $\phi(y+mx)$; repeated root gives an extra factor $x$.
• P.I. for $e^{ax+by}$: $1/F(a,b)$; for trig: $D^2\to-a^2$, $DD'\to-ab$, $D'^2\to-b^2$.
• Answers: $\frac{e^{x+2y}}{27}$, $-\frac{\cos(x-2y)}{25}$, $\frac{\sin(2x+3y)}{17}$, $10x^4+6x^3y$, $\frac{x^4y}{12}$.

Memory hooks

• "Lagrange = Ladder of ratios": three fractions, two integrals, one $f(u,v)=0$.
• "Charpit needs five derivatives": $f_x,f_y,f_z,f_p,f_q$.
• "Root $m$ pairs with $y+mx$": repeated root, extra $x$.
• "Trig: square the letters": $D^2\to-a^2$, $D'^2\to-b^2$, $DD'\to-ab$.

Coverage checklist

• Formulation of Partial Differential equations: $z=f(y/x)$; $z=y^2+2f(1/x+\log y)$; $f(x^2+y^2+z^2,z^2-2xy)=0$.
• Linear and Non-Linear Partial Differential Equations: $(x-y)p+(x+y)q=2xz$; $(y+z)p+(x+z)q=x+y$; $(p^2+q^2)y=qz$; $x^2p^2+y^2q^2=z^2$; $px+qy=pq$.
• Homogeneous Linear Partial Differential Equations with Constants Coefficients: $e^{x+2y}$; $\cos(x-2y)$; $\sin(2x+3y)$; $\cos(2x+y)$; $12x^2+36xy$; $z_{xx}-z_{yy}=x^2y$.