Notes - Mathematics Olympiads -Differential Equations
Category : 12th Class
Differential Equations
e.g. An equation of the form of
\[\therefore \,\,\,\,\,y=f(x,y,p)\]is said to be differential equation, where\[p=\frac{dy}{dx}\]e.g.
(1) \[y={{\left( \frac{dy}{dx} \right)}^{3}}+5\] (2) \[{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}+y\frac{dy}{dx}=5y+c\] etc.
The order of highest order derivative occurred in differential equation is said to be the order of the differential equation whereas power/exponent of the highest order derivative term in the different equation whereas power/exponent of the highest order derivative term in the differential equation is said to be the degree of the differential equation.
e.g.
(1) \[\frac{{{d}^{3}}y}{d{{x}^{3}}}+2{{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}^{2}}+6{{\left( \frac{dy}{dx} \right)}^{6}}+7y=0\]
Here order of differential equation be 3
Degree of differential equation = 1
(2) \[{{\left[ 1+{{\left( \frac{dy}{dx} \right)}^{2}} \right]}^{\frac{3}{2}}}=\frac{{{d}^{3}}y}{d{{x}^{3}}}\]
Order of this differential equation = 3
Degree of this differential equation = 2
Types of Differential Equation:
If \[u=f(x,y,z),\] then its partial differential equation (P.D.E) will be
\[\frac{{{\partial }^{2}}u}{\partial {{x}^{2}}}+\frac{{{\partial }^{2}}u}{\partial {{y}^{2}}}+\frac{{{\partial }^{2}}u}{\partial {{z}^{2}}}=0\]
An equation of the form \[\frac{dy}{dx}=\frac{M}{N},\]where M and N be the functions of x and y or any constant value, is said to be the first order, first degree differential equation.
This type of differential equation consists of following types of function
Solved Example
Solution: \[\frac{dy}{dx}=(1+{{x}^{2}}).(1+{{y}^{2}})\]
\[\Rightarrow \] \[\frac{dy}{1+{{y}^{2}}}=(1+{{x}^{2}})dx\]
Integrating both sides, we get
\[{{\tan }^{-1}}y=x+\frac{{{x}^{3}}}{2}+c\]
\[\Rightarrow \] \[y=\tan \left( \frac{2x+{{x}^{3}}+2c}{2} \right)\]
Step 1: First of all arrange the given equation as \[\frac{dy}{dx}=\frac{\phi (x)}{f(x)},\] where \[\phi (x)\] and \[f(x)\] having same degree.
Step2: Put\[y=\text{v}x\]and differentiate it on both sides to get \[\frac{dy}{dx}=\text{v}+x\frac{d\text{v}}{dx}\]
Step 3: Then put these value in the given equation and further hence using variable seperable method, solution will be obtained in the form of v.
Step 4: Then replace v by \[\frac{y}{x}\] & hence find the result.
Solved Example
Solution:\[y'=\frac{dx}{dx}=\frac{x+y}{x}.\] (1)
It is a homogeneous equation.
Now, put \[y=\text{v}x\] (2)
On differentiating both sides of equ (2), we get
\[\frac{dy}{dx}=\text{v}+x\frac{d\text{v}}{dx}\]
On replacing the value of \[\frac{dy}{dx}\] in equ (1), it becomes
\[\text{v}+x\frac{d\text{v}}{dx}=\frac{x+\text{v}x}{x}=(1+\text{v})\]
\[\Rightarrow \,\,\,x\frac{d.\text{v}}{dx}=1+\text{v}\,-\,\text{v=1}\] \[\Rightarrow \] \[d\text{v}=\frac{dx}{x}\]
On integrating both sides, we get
\[\text{v}=\log x+\log c\]
\[\frac{y}{x}=\log xc\](where log \[c=\] integration constant) \[y=x\log xc\]
which is the required solution.
Step 1: First of all, find the integrating factor (I.F.)
I.F. \[={{e}^{\int P.dx}}\]
Step 2: Then, the general solution will be given as \[y(I.F.)\]\[=\int{Q.(I.F.)dx}\]
Solved Example
Solution: Given \[\frac{dy}{dx}+\left( \frac{1}{x} \right).y={{x}^{3}}\]
which is a Linear Differential Equation (L.D.E.)
Here, \[P=\frac{1}{x}\]and\[Q={{x}^{3}}\]
Now I.F\[={{e}^{\int P.dx}}\]\[={{e}^{\int \frac{1}{x}.dx}}\]\[={{e}^{log\,x}}\]\[=x\]
\[\therefore \] Hence the general solution will be
\[y.(I.F)=\int{Q.(I.F).dx}\]
\[y.x=\int{{{x}^{3}}}.x.dx=\int{{{x}^{4}}.dx}\]
\[\Rightarrow yx=\frac{{{x}^{5}}}{5}+c\]
\[\Rightarrow 5xy={{x}^{5}}+K\]
which is the required solution.
where P and Q is in the form of y or any constant (but not in x)
For solving this type of LDE, we adopt the same procedures as in the previous case.
Step 1: First of all, find I.F. (Intergrating factor)
I.F.\[={{e}^{\int Pdy}}\]
Step 2: Further hence, the general solution will be given by
\[x.I.F=\int{Q.(I.F.)dy}\]
\[\frac{dy}{dx}+Py=Q.{{y}^{n}}\] (1)
Where P and Q are the functions of x, which do not appear directly to be of linear form, but can be easily reduced into the linear form by a suitable transformation.
Step 1: Divide both sides of equation (1) by\[{{y}^{n}}\], we get
\[\frac{1}{{{y}^{n}}}\frac{dy}{dx}+P.\frac{1}{{{y}^{n-1}}}=Q\] (2)
Step 2: Then put \[\frac{1}{{{y}^{n-1}}}={{y}^{-n+1}}=z\]
\[\Rightarrow \ \ \ \frac{dz}{dx}=(-n+1).{{y}^{-n}}.\frac{dy}{dx}\] \[\Rightarrow \] \[\frac{(1-n)}{{{y}^{n}}}.\frac{dy}{dx}=\frac{dz}{dx}\]
Then equation (2) becomes
\[\frac{1}{1-n}.\frac{dz}{dx}+Pz=Q\] \[\Rightarrow \] \[\frac{dz}{dx}+(1-n).\]\[Pz=(1-n).\]Q
This equation is linear in z. And hence it can be solved by the methods discussed in previous section.
Solved Example
Sol. Dividing both sides by \[{{y}^{2}}\], we get
\[\frac{1}{{{y}^{2}}}.\frac{dy}{dx}+\frac{1}{xy}=\frac{1}{{{x}^{2}}}\] (1)
Now put \[\frac{1}{y}=z,\] it gives
\[\frac{dz}{dx}=\frac{-1}{{{y}^{2}}}.\frac{dy}{dx}\]
Now the equation (1) reduces to
\[\frac{-dz}{dx}+\frac{z}{x}=\frac{1}{{{x}^{2}}}\] \[\Rightarrow \frac{dz}{dx}-\frac{z}{x}=\frac{-1}{{{x}^{2}}}\]
which is the linear equation in the form of z.
Here, \[P=\frac{-1}{x}\]and \[Q=\frac{1}{{{x}^{2}}}\]
I.F. \[={{e}^{\int \frac{-1}{x}.dx}}\] \[={{e}^{-\log x}}={{e}^{\log \frac{1}{x}}}=\frac{1}{x}\]
Hence the general solution will be
\[z.\frac{1}{x}=\int{\frac{-1}{{{x}^{2}}}.\frac{1}{x}.dx}\]
\[\Rightarrow \frac{z}{x}=\int{\frac{-1}{{{x}^{3}}}.dx}=\frac{-1}{2}.\frac{1}{{{x}^{2}}}+c\]
\[\Rightarrow \frac{z}{x}=\frac{1}{2{{x}^{2}}}+c\]\[\Rightarrow \frac{1}{yx}=\frac{1}{2{{x}^{2}}}+c\]
\[\Rightarrow 2x=y+2{{x}^{2}}yc\]\[\Rightarrow y=\frac{2x}{1+k{{x}^{2}}}\]
which is the required solution.
Sol. Given Differential Equation is
\[x\frac{dy}{dx}+2y={{x}^{2}}\log x\]
Dividing by x on both sides, we get
\[\frac{dy}{dx}+\frac{2y}{x}=x\log x\]
Here, \[P=\frac{2}{x}\]and \[Q=x\log x\]
\[\therefore \] Integrating factor, I.F.\[={{e}^{\int P.dx}}\]
\[={{e}^{\int \frac{2}{x}.dx}}\]\[={{e}^{\log {{x}^{2}}}}\]\[={{x}^{2}}\]
Hence the general solution will be
\[y(I.F.)=\int{Q.(I.F.)dx}\]
\[\Rightarrow \]\[y{{x}^{2}}=\int{(x\log x){{x}^{2}}.dx}\] \[\Rightarrow y{{x}^{2}}=\int{{{x}^{3}}.\log x.dx}\]
\[\Rightarrow y{{x}^{2}}=\log x\int{{{x}^{3}}dx-\int{\left( \frac{d(\log x)}{dx}\int{{{x}^{3}}dx} \right)dx}}\]
(Integrating by parts)
\[\Rightarrow y{{x}^{2}}=\frac{{{x}^{4}}}{4}\log x-\int{\left( \frac{1}{x}.\frac{{{x}^{4}}}{4} \right)dx}\] \[\Rightarrow y{{x}^{2}}=\frac{{{x}^{4}}}{4}\log x-\int{\frac{{{x}^{3}}}{4}dx}\]
\[\Rightarrow y{{x}^{2}}=\frac{{{x}^{4}}}{4}\log x-\frac{{{x}^{4}}}{16}+c\] \[\Rightarrow y=\frac{{{x}^{2}}\log x}{4}-\frac{{{x}^{2}}}{16}+c\]
\[\Rightarrow y=\frac{{{x}^{2}}}{16}(4\log x-1)+c\]
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