अवकल समीकरण
हल की विधि
सामान्य हल
चर अलग करने योग्य समीकरण (Separable equations)
First-order, separable in x and y (general case, see below for special cases)[1]
P
1
(
x
)
Q
1
(
y
)
+
P
2
(
x
)
Q
2
(
y
)
d
y
d
x
=
0
{\displaystyle P_{1}(x)Q_{1}(y)+P_{2}(x)Q_{2}(y)\,{\frac {dy}{dx}}=0\,\!}
P
1
(
x
)
Q
1
(
y
)
d
x
+
P
2
(
x
)
Q
2
(
y
)
d
y
=
0
{\displaystyle P_{1}(x)Q_{1}(y)\,dx+P_{2}(x)Q_{2}(y)\,dy=0\,\!}
Separation of variables (divide by P 2 Q 1 ).
∫
x
P
1
(
λ
)
P
2
(
λ
)
d
λ
+
∫
y
Q
2
(
λ
)
Q
1
(
λ
)
d
λ
=
C
{\displaystyle \int ^{x}{\frac {P_{1}(\lambda )}{P_{2}(\lambda )}}\,d\lambda +\int ^{y}{\frac {Q_{2}(\lambda )}{Q_{1}(\lambda )}}\,d\lambda =C\,\!}
First-order, separable in x [2]
d
y
d
x
=
F
(
x
)
{\displaystyle {\frac {dy}{dx}}=F(x)\,\!}
d
y
=
F
(
x
)
d
x
{\displaystyle dy=F(x)\,dx\,\!}
Direct integration.
y
=
∫
x
F
(
λ
)
d
λ
+
C
{\displaystyle y=\int ^{x}F(\lambda )\,d\lambda +C\,\!}
First-order, autonomous, separable in y [2]
d
y
d
x
=
F
(
y
)
{\displaystyle {\frac {dy}{dx}}=F(y)\,\!}
d
y
=
F
(
y
)
d
x
{\displaystyle dy=F(y)\,dx\,\!}
Separation of variables (divide by F ).
x
=
∫
y
d
λ
F
(
λ
)
+
C
{\displaystyle x=\int ^{y}{\frac {d\lambda }{F(\lambda )}}+C\,\!}
First-order, separable in x and y [2]
P
(
y
)
d
y
d
x
+
Q
(
x
)
=
0
{\displaystyle P(y){\frac {dy}{dx}}+Q(x)=0\,\!}
P
(
y
)
d
y
+
Q
(
x
)
d
x
=
0
{\displaystyle P(y)\,dy+Q(x)\,dx=0\,\!}
Integrate throughout.
∫
y
P
(
λ
)
d
λ
+
∫
x
Q
(
λ
)
d
λ
=
C
{\displaystyle \int ^{y}P(\lambda )\,{d\lambda }+\int ^{x}Q(\lambda )\,d\lambda =C\,\!}
General first-order equations
First-order, homogeneous[2]
d
y
d
x
=
F
(
y
x
)
{\displaystyle {\frac {dy}{dx}}=F\left({\frac {y}{x}}\right)\,\!}
Set y = ux , then solve by separation of variables in u and x .
ln
(
C
x
)
=
∫
y
/
x
d
λ
F
(
λ
)
−
λ
{\displaystyle \ln(Cx)=\int ^{y/x}{\frac {d\lambda }{F(\lambda )-\lambda }}\,\!}
First-order, separable[1]
y
M
(
x
y
)
+
x
N
(
x
y
)
d
y
d
x
=
0
{\displaystyle yM(xy)+xN(xy)\,{\frac {dy}{dx}}=0\,\!}
y
M
(
x
y
)
d
x
+
x
N
(
x
y
)
d
y
=
0
{\displaystyle yM(xy)\,dx+xN(xy)\,dy=0\,\!}
Separation of variables (divide by xy ).
ln
(
C
x
)
=
∫
x
y
N
(
λ
)
d
λ
λ
[
N
(
λ
)
−
M
(
λ
)
]
{\displaystyle \ln(Cx)=\int ^{xy}{\frac {N(\lambda )\,d\lambda }{\lambda [N(\lambda )-M(\lambda )]}}\,\!}
If N = M , the solution is xy = C .
Exact differential , first-order[2]
M
(
x
,
y
)
d
y
d
x
+
N
(
x
,
y
)
=
0
{\displaystyle M(x,y){\frac {dy}{dx}}+N(x,y)=0\,\!}
M
(
x
,
y
)
d
y
+
N
(
x
,
y
)
d
x
=
0
{\displaystyle M(x,y)\,dy+N(x,y)\,dx=0\,\!}
where
∂
M
∂
x
=
∂
N
∂
y
{\displaystyle {\frac {\partial M}{\partial x}}={\frac {\partial N}{\partial y}}\,\!}
Integrate throughout.
F
(
x
,
y
)
=
∫
y
M
(
x
,
λ
)
d
λ
+
∫
x
N
(
λ
,
y
)
d
λ
+
Y
(
y
)
+
X
(
x
)
=
C
{\displaystyle {\begin{aligned}F(x,y)&=\int ^{y}M(x,\lambda )\,d\lambda +\int ^{x}N(\lambda ,y)\,d\lambda \\&+Y(y)+X(x)=C\end{aligned}}\,\!}
where Y (y ) and X (x ) are functions from the integrals rather than constant values, which are set to make the final function F (x, y ) satisfy the initial equation.
Inexact differential , first-order[2]
M
(
x
,
y
)
d
y
d
x
+
N
(
x
,
y
)
=
0
{\displaystyle M(x,y){\frac {dy}{dx}}+N(x,y)=0\,\!}
M
(
x
,
y
)
d
y
+
N
(
x
,
y
)
d
x
=
0
{\displaystyle M(x,y)\,dy+N(x,y)\,dx=0\,\!}
where
∂
M
∂
x
≠
∂
N
∂
y
{\displaystyle {\frac {\partial M}{\partial x}}\neq {\frac {\partial N}{\partial y}}\,\!}
Integration factor μ(x, y ) satisfying
∂
(
μ
M
)
∂
x
=
∂
(
μ
N
)
∂
y
{\displaystyle {\frac {\partial (\mu M)}{\partial x}}={\frac {\partial (\mu N)}{\partial y}}\,\!}
If μ(x, y ) can be found:
F
(
x
,
y
)
=
∫
y
μ
(
x
,
λ
)
M
(
x
,
λ
)
d
λ
+
∫
x
μ
(
λ
,
y
)
N
(
λ
,
y
)
d
λ
+
Y
(
y
)
+
X
(
x
)
=
C
{\displaystyle {\begin{aligned}F(x,y)&=\int ^{y}\mu (x,\lambda )M(x,\lambda )\,d\lambda +\int ^{x}\mu (\lambda ,y)N(\lambda ,y)\,d\lambda \\&+Y(y)+X(x)=C\\\end{aligned}}\,\!}
General second-order equations
Second-order, autonomous[3]
d
2
y
d
x
2
=
F
(
y
)
{\displaystyle {\frac {d^{2}y}{dx^{2}}}=F(y)\,\!}
Multiply equation by 2dy/dx , substitute
2
d
y
d
x
d
2
y
d
x
2
=
d
d
x
(
d
y
d
x
)
2
{\displaystyle 2{\frac {dy}{dx}}{\frac {d^{2}y}{dx^{2}}}={\frac {d}{dx}}\left({\frac {dy}{dx}}\right)^{2}\,\!}
, then integrate twice.
x
=
±
∫
y
d
λ
2
∫
λ
F
(
ϵ
)
d
ϵ
+
C
1
+
C
2
{\displaystyle x=\pm \int ^{y}{\frac {d\lambda }{\sqrt {2\int ^{\lambda }F(\epsilon )\,d\epsilon +C_{1}}}}+C_{2}\,\!}
Linear equations (up to n th order)
First-order, linear, inhomogeneous, function coefficients[2]
d
y
d
x
+
P
(
x
)
y
=
Q
(
x
)
{\displaystyle {\frac {dy}{dx}}+P(x)y=Q(x)\,\!}
Integrating factor:
e
∫
x
P
(
λ
)
d
λ
{\displaystyle e^{\int ^{x}P(\lambda )\,d\lambda }}
.
y
=
e
−
∫
x
P
(
λ
)
d
λ
[
∫
x
e
∫
λ
P
(
ϵ
)
d
ϵ
Q
(
λ
)
d
λ
+
C
]
{\displaystyle y=e^{-\int ^{x}P(\lambda )\,d\lambda }\left[\int ^{x}e^{\int ^{\lambda }P(\epsilon )\,d\epsilon }Q(\lambda )\,{d\lambda }+C\right]}
Second-order, linear, inhomogeneous, constant coefficients[4]
d
2
y
d
x
2
+
b
d
y
d
x
+
c
y
=
r
(
x
)
{\displaystyle {\frac {d^{2}y}{dx^{2}}}+b{\frac {dy}{dx}}+cy=r(x)\,\!}
Complementary function yc : assume yc = e αx , substitute and solve polynomial in α, to find the linearly independent functions
e
α
j
x
{\displaystyle e^{\alpha _{j}x}}
.
Particular integral yp : in general the method of variation of parameters , though for very simple r (x ) inspection may work.[2]
y
=
y
c
+
y
p
{\displaystyle y=y_{c}+y_{p}}
If b 2 > 4c , then:
y
c
=
C
1
e
(
−
b
+
b
2
−
4
c
)
x
2
+
C
2
e
−
(
b
+
b
2
−
4
c
)
x
2
{\displaystyle y_{c}=C_{1}e^{\left(-b+{\sqrt {b^{2}-4c}}\right){\frac {x}{2}}}+C_{2}e^{-\left(b+{\sqrt {b^{2}-4c}}\right){\frac {x}{2}}}\,\!}
If b 2 = 4c , then:
y
c
=
(
C
1
x
+
C
2
)
e
−
b
x
/
2
{\displaystyle y_{c}=(C_{1}x+C_{2})e^{-bx/2}\,\!}
If b 2 < 4c , then:
y
c
=
e
−
b
x
2
[
C
1
sin
(
|
b
2
−
4
c
|
x
2
)
+
C
2
cos
(
|
b
2
−
4
c
|
x
2
)
]
{\displaystyle y_{c}=e^{-b{\frac {x}{2}}}\left[C_{1}\sin {\left({\sqrt {\left|b^{2}-4c\right|}}{\frac {x}{2}}\right)}+C_{2}\cos {\left({\sqrt {\left|b^{2}-4c\right|}}{\frac {x}{2}}\right)}\right]\,\!}
n th-order, linear, inhomogeneous, constant coefficients[4]
∑
j
=
0
n
b
j
d
j
y
d
x
j
=
r
(
x
)
{\displaystyle \sum _{j=0}^{n}b_{j}{\frac {d^{j}y}{dx^{j}}}=r(x)\,\!}
Complementary function yc : assume yc = e αx , substitute and solve polynomial in α, to find the linearly independent functions
e
α
j
x
{\displaystyle e^{\alpha _{j}x}}
.
Particular integral yp : in general the method of variation of parameters , though for very simple r (x ) inspection may work.[2]
y
=
y
c
+
y
p
{\displaystyle y=y_{c}+y_{p}}
Since αj are the solutions of the polynomial of degree n :
∏
j
=
1
n
(
α
−
α
j
)
=
0
{\displaystyle \prod _{j=1}^{n}\left(\alpha -\alpha _{j}\right)=0\,\!}
, then:
for αj all different,
y
c
=
∑
j
=
1
n
C
j
e
α
j
x
{\displaystyle y_{c}=\sum _{j=1}^{n}C_{j}e^{\alpha _{j}x}\,\!}
for each root αj repeated kj times,
y
c
=
∑
j
=
1
n
(
∑
ℓ
=
1
k
j
C
ℓ
x
ℓ
−
1
)
e
α
j
x
{\displaystyle y_{c}=\sum _{j=1}^{n}\left(\sum _{\ell =1}^{k_{j}}C_{\ell }x^{\ell -1}\right)e^{\alpha _{j}x}\,\!}
for some αj complex, then setting α = χj + i γj , and using Euler's formula , allows some terms in the previous results to be written in the form
C
j
e
α
j
x
=
C
j
e
χ
j
x
cos
(
γ
j
x
+
ϕ
j
)
{\displaystyle C_{j}e^{\alpha _{j}x}=C_{j}e^{\chi _{j}x}\cos(\gamma _{j}x+\phi _{j})\,\!}
where ϕj is an arbitrary constant (phase shift).