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Definition Of Differential Equations And Their Solutions

Learn definition of a Differential Equation & order of differential equations, Practice to find particular solution of the differential equation problems.

 Differential Equation and Its Order

  • An equation that contains independent variable (usually \('x'\)) some functions of \(x\), dependent variable (usually \('y'\)) some functions of \(y\) and at least one derivative term of \(y\) with respect to  \(x\) is called a differential equation.

e.g (1)  \(\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}=sin\,x\)

(2)  \(\dfrac{2ydy}{dx}+lnx=sin^2\,x\)

(3)  \({\sqrt{1+\dfrac {dy}{dx}}}=\dfrac{d^2y}{dx^2}\) are all differential equations.

  • The order of differential equations is the order of the highest order derivative that occurs in the equations.

Example: (1)  \(\dfrac{d^2y}{dx^2}+\dfrac{xdy}{dx}=5sin\,x\) is of order 2. \(\left(\because\,\dfrac{d^2y}{dx^2}\text{represents order 2}\right)\)

(2)  \(x^3\dfrac{dy}{dx}+lnx=5\) is of order 1. \(\left(\because\,\dfrac{dy}{dx}\text{represents order 1}\right)\)

(3)  \(5y^3\dfrac{d^3y}{dx^3}=2x+\dfrac{dy}{dx}=5sin\,x\) is of order 3. \(\left(\because\,\dfrac{d^3y}{dx^3}\text{represents order 3}\right)\)

Illustration Questions

Which of the following equations is a differential equation?

A \(2y^3–x^2=sin\,x\)

B \(cos^3x+e^4\,lnx=2\)

C \(5y+x=7\)

D \(\dfrac{xdy}{dx}+y=12cos^3x\)

×

Differential equations should contain at least one derivative term, other than \('x'\) and \('y'\) terms.

Options 'A', 'B' and 'C' do not contain any derivative term, while 'd' has a \(\dfrac{dy}{dx}\) term 

\(\therefore\) Option 'D' is correct.

Which of the following equations is a differential equation?

A

\(2y^3–x^2=sin\,x\)

.

B

\(cos^3x+e^4\,lnx=2\)

C

\(5y+x=7\)

D

\(\dfrac{xdy}{dx}+y=12cos^3x\)

Option D is Correct

Illustration Questions

The order of the differential equation  \(5y^2\dfrac{dy}{dx}+2sin^2x=\dfrac{d^2y}{dx^2}\) is

A 1

B 2

C 3

D 4

×

The order of differential equation is the order of highest order derivative that occurs in the equations.

\(\dfrac{d^2y}{dx^2}\) is the highest order derivative in the equations.

\(\therefore\) Order of differential equation is 2

The order of the differential equation  \(5y^2\dfrac{dy}{dx}+2sin^2x=\dfrac{d^2y}{dx^2}\) is

A

1

.

B

2

C

3

D

4

Option B is Correct

Solution of a Differential Equation

  • A function \(f\) is called solution to a differential equation if the equation is satisfied when \(y=f(x)\) and its derivative are substituted in the equation.

\(f(x)\) is a solution to \(y'=5y\) if

\(f'(x)=5f(x)\)

  • \(f(x)=x^2+2x\) is a solution to \(y'=2x+2\)
  • one differential equation may have more than one solution that's why we have to check all the options one by one rather than finding the particular solution.

Illustration Questions

Which of the following is a solution to the differential equation \(\dfrac{d^2y}{dx^2}=\dfrac{1}{y}\left(\dfrac{dy}{dx}\right)^2\)?

A \(f(x)=y=sin^2x\)

B \(f(x)=y=cos^2x\)

C \(f(x)=y=2e^{3x}\)

D \(f(x)=y=ln2x\)

×

\(y=f(x)\) is a solution to a differential equation if the equation is satisfied when \(y=f(x)\) and its derivative is substituted in the equation.

For option (A) 

\(y=sin^2x\Rightarrow\,\dfrac{dy}{dx}=2sinx\,cosx=sin2x\) & \(\dfrac{d^2y}{dx^2}=2cos2x\)

substituting in the differential equation, we have

\(2cos2x=\dfrac{1}{sin^2x}×sin^22x\to\)not true

For option (B) 

\(y=cos^2x\Rightarrow\,\dfrac{dy}{dx}=-2cosx\,sinx=-sin2x\) & \(\dfrac{d^2y}{dx^2}=-2cos2x\)

substituting in the differential equation, we have

\(-2cos2x=\dfrac{1}{cos^2x}×sin^22x\to\)not true

For option (C) 

\(y=2e^{3x}\Rightarrow\,\dfrac{dy}{dx}=6e^{3x} \,\& \,\dfrac{d^2y}{dx^2}=18e^{3x}\)

substituting in the differential equation, we have

\(18e^{3x}=\dfrac{1}{2e^{3x}}×36e^{6x}\to18e^{3x}=18e^{3x}\Rightarrow\)  true

For option (D) 

\(y=ln2x\Rightarrow\,\dfrac{dy}{dx}=\dfrac{1}{2x}×2=\dfrac{1}{x}\) & \(\dfrac{d^2y}{dx^2}=\dfrac{-1}{x^2}\)

substituting in the differential equation, we have

\(\dfrac{-1}{x^2}=\dfrac{1}{ln2x}×\dfrac{1}{x^2}\to\)not true

\(\therefore\) Option (C) is correct.

Which of the following is a solution to the differential equation \(\dfrac{d^2y}{dx^2}=\dfrac{1}{y}\left(\dfrac{dy}{dx}\right)^2\)?

A

\(f(x)=y=sin^2x\)

.

B

\(f(x)=y=cos^2x\)

C

\(f(x)=y=2e^{3x}\)

D

\(f(x)=y=ln2x\)

Option C is Correct

Finding the Value of a Parameter of Differential Equation for the Given Solution

  •  A function \(f\) is called solution to a differential equation if the equation is satisfied when \(y=f(x)\) and its derivative are substituted in the equation.
  • Suppose we desire to know the value of \(\alpha\) for which \(y=e^{\alpha x}\) is a solution to the differential equation.

    \(y''+2y'+y=0\)

    Find \(y'=\alpha e^{\alpha x}\) and \(y''=\alpha ^2e^{\alpha x}\) and put in the equation.

    \(\alpha^2 e^{\alpha x}+2\alpha e^{\alpha x}+e^{\alpha x}=0\)

    \(\Rightarrow \,e^{\alpha x}(\alpha^2+2\alpha+1)=0\)

    \(\Rightarrow\,e^{\alpha x}(\alpha+1)^2=0\)

    \(\Rightarrow\,\alpha=–1\)

     

     

 

Illustration Questions

If \(y=e^{\alpha x}\) is a solution to the differential equation \(y''+5y'+6y=0\). Then the possible value(s) of constant  \('\alpha'\) are   

A \(\alpha=-2,\,\alpha=-3\)

B \(\alpha=-5,\,\alpha=\dfrac{1}{2}\)

C \(\alpha=1,\,\alpha=6\)

D \(\alpha=4,\,\alpha=1\)

×

\(y=f(x)\) is a solution to a differential equation if the equation is satisfied when \(y=f(x)\) and its derivatives are substituted in the equation.

In this case :

\(y=e^{\alpha x}\Rightarrow\,\dfrac{dy}{dx}=y'=\alpha e^{\alpha x}\Rightarrow\,\dfrac{d^2y}{dx^2}=y''=\alpha^2e^{\alpha x}\)

 

Putting the above in the differential equation 

\(\alpha^2e^{\alpha x}+5\alpha e^{\alpha x}+6 e^{\alpha x}=0\)

\(\Rightarrow\,e^{\alpha x} (\alpha^2+5\alpha+6)=0\)  (we know \(e^{\alpha x}\ne\,0\) for any \(x\)).

\(\therefore\,\alpha^2+5\alpha+6=0\Rightarrow\,(\alpha+2)(\alpha+3)=0\Rightarrow\,\alpha=-2,\,\alpha=-3\)

If \(y=e^{\alpha x}\) is a solution to the differential equation \(y''+5y'+6y=0\). Then the possible value(s) of constant  \('\alpha'\) are   

A

\(\alpha=-2,\,\alpha=-3\)

.

B

\(\alpha=-5,\,\alpha=\dfrac{1}{2}\)

C

\(\alpha=1,\,\alpha=6\)

D

\(\alpha=4,\,\alpha=1\)

Option A is Correct

Illustration Questions

If \(y=sin\,kx\) is a solution to the differential equation \(36y''=–49y\). Then the value of \('k'\) is  

A \(k=\pm\dfrac{5}{4}\)

B \(k=\pm{3}\)

C \(k=\pm{9}\)

D \(k=\pm\dfrac{7}{6}\)

×

\(y=f(x)\) is a solution to a differential equation if the equation is satisfied when \(y=f(x)\) and its derivatives are substituted in the equation.

In this case :

\(y=sin\,kx\Rightarrow y'=\,\dfrac{dy}{dx}=kcos\,kx \Rightarrow y''=\,\dfrac{d^2y}{dx^2}\)

\(=–k^2sin\,kx\)

Putting the above in the differential equation have

\(36×–k^2sin\,kx=–49sin\,kx\)

\(36k^2=49\Rightarrow\,k=\pm\dfrac{7}{6}\Rightarrow\,k=\dfrac{7}{6}\) or \(k=–\dfrac{7}{6}\)

If \(y=sin\,kx\) is a solution to the differential equation \(36y''=–49y\). Then the value of \('k'\) is  

A

\(k=\pm\dfrac{5}{4}\)

.

B

\(k=\pm{3}\)

C

\(k=\pm{9}\)

D

\(k=\pm\dfrac{7}{6}\)

Option D is Correct

General and Particular Solution

When the solution to a differential equation is asked usually all solutions have to be reported

  • Consider the differential equation 

\(f'(x)=f(x)\to\) we know that the solution to this equation is \(f(x)=e^x\) but all function of the form \(f(x)=e^x\) are also have solutions we say that \(y=ce^x\) is called the general solution (or family of solution) to the differential equation.

  • sometimes we are not interested in finding general solution but a solution that satisfies some particular condition, such a solution is called particular solution to the differential equation. The condition is called initial condition and the problem is called initial value problem (IVP).

e.g.  \(y'=y\to y(0)=1^{ \nearrow^{\text {initial condition}}}\)

\(\Rightarrow\,y=ce^x \to\) general solution 

Use initial condition

\(1=ce^0\Rightarrow\,c=1\)

\(\therefore\,y=e^x \to\) particular solution.

Illustration Questions

If \(y=xe^x+ce^x\) is the general solution to the differential equation \(y'-y=e^x\), find the solution to the initial value problem \(y(0)=1\).

A \(y=2sinx+lnx\)

B \(y=e^x(x+1)\)

C \(y=ce^x+4\) (C \(\in\)R)

D \(y=e^x(x^2+1)\)

×

Find 'c' using the initial condition and then put this of c in the general solution to get particular solution.

General solution is \(y=xe^x+ce^x\)

Use \(y(0)=1\Rightarrow\,1=0e^0+ce^0\Rightarrow1=0+c\Rightarrow c=1\)

\(\therefore\,\) The particular solution or solution to initial value problem is

\(y=xe^x+e^x\Rightarrow\,y=e^x(x+1)\)

If \(y=xe^x+ce^x\) is the general solution to the differential equation \(y'-y=e^x\), find the solution to the initial value problem \(y(0)=1\).

A

\(y=2sinx+lnx\)

.

B

\(y=e^x(x+1)\)

C

\(y=ce^x+4\) (C \(\in\)R)

D

\(y=e^x(x^2+1)\)

Option B is Correct

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