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Variable Separation Method

Find the solution of differential equation by variable separable method, Practice differential equation of a curve by variable separable method & advanced application.

Standard Form of Variable Separable Differential Equation and Its Solution

A differential equation of the form

\(\dfrac{dy}{dx}=f(x,\,y)\) is called variable separable

if \(f(x,\,y)\) is of the form \(f(x,\,y)=g(x)\,h(y)\)

We will now find the solution to a particular category of differential equation where \(y\) is expressed explicitly as a function of \(x\).

  • Consider an equation of the form

\(\dfrac{dy}{dx}=f(x,\,y)\;\to\) (1)

Such that \(f(x,\,y)=g(x)\;h(y)\). i.e. R.H.S. expression can be expressed as function of \(x\) times function of \(y\).

\(\therefore\) differential equation (1) can be written as

\(\dfrac{dy}{dx}=g(x)\;h(y)\)

  • The name variable separable method, comes from the fact that terms of \(x\) and \(y\) can be separated in R.H.S.
  • To solve this differential equation we write it as:

\(\dfrac{dy}{h(y)}=g(x)\,dx\) (all \(x\) terms with \(dx\), all \(y\) terms with \(dy\))

Now we integrate both sides of the equation

\(\displaystyle\int \dfrac{dy}{h(y)}=\int g(x)\,dx\)

Add  the integration constant  \(C\)  on any one side (R.H.S. or L.H.S.)

  • The relation obtained between \(y\) and \(x\) in called the general solution of differential equation.

After obtaining the general solution we can find the value of C by using an initial condition given in the problem. This initial condition is usually the value of \(y\) at a particular value of \(x\).

Illustration Questions

Which one of the following differential equations is variable separable ?

A \(\dfrac{dy}{dx}=2x+y\)

B \(\dfrac{dy}{dx}=x\,sin\,y\)

C \(\dfrac{dy}{dx}+sin^2x=cos^2y\)

D \(x\dfrac{dy}{dx}=ln\,x+y\)

×

A differential equation of the form

\(\dfrac{dy}{dx}=f(x,\,y)\) is called variable separable if \(f(x,\,y)\) is of the form \(f(x,\,y)=g(x)\;h(y)\).

Out of the 4 options, only B can be expressed in the above form

\(\dfrac{dy}{dx}=x\,sin\,y\)

\(\Rightarrow\;\dfrac{dy}{sin\,y}=x\,dx\)

\(\therefore\)  Option (B) is correct.

Which one of the following differential equations is variable separable ?

A

\(\dfrac{dy}{dx}=2x+y\)

.

B

\(\dfrac{dy}{dx}=x\,sin\,y\)

C

\(\dfrac{dy}{dx}+sin^2x=cos^2y\)

D

\(x\dfrac{dy}{dx}=ln\,x+y\)

Option B is Correct

Illustration Questions

Solve the differential equation \(5\dfrac{dy}{dx}=e^xy^4\)

A \(e^x=y^4+C\)

B \(sin\,x=y\,cos\,y+C\)

C \(y^2=5e^x+C\)

D \(y^3=\dfrac{-5}{3(e^x+C)}\)

×

For solving equations of the form

\(\dfrac{dy}{dx}=h(x)\,g(y)\) separate the variables and integrate \(\to\;\displaystyle \int\dfrac{dy}{g(y)}=\int h(x)\,dx\)

In this case 

\(5\dfrac{dy}{dx}=e^xy^4\)

\(\Rightarrow\;\dfrac{dy}{dx}=\dfrac{e^xy^4}{5}\)

\(\Rightarrow\;\dfrac{5}{y^4}dy=e^xdx\)

\(\Rightarrow\;5y^{-4}dy=e^xdx\)

Integrate both sides 

\(\displaystyle\int 5y^{-4}dy=\displaystyle\int e^xdx\)

\(\Rightarrow\;\dfrac{5y^{-3}}{-3}=e^x+C\)

\(\Rightarrow\;\dfrac{-5}{3y^3}=e^x+C\)

\(\Rightarrow\;y^3=\dfrac{-5}{3(e^x+C)}\)

Solve the differential equation \(5\dfrac{dy}{dx}=e^xy^4\)

A

\(e^x=y^4+C\)

.

B

\(sin\,x=y\,cos\,y+C\)

C

\(y^2=5e^x+C\)

D

\(y^3=\dfrac{-5}{3(e^x+C)}\)

Option D is Correct

Illustration Questions

Solve the differential equation \(\dfrac{dy}{dx}=e^{x+y}+e^yx^3\) and \(y\) in terms of \(x\) in the solution.

A \(y=ln\left(e^x+x^2-x\right)+C\)

B \(y=2sin^2x-cos\,x+C\)

C \(y=-ln\left(-e^x-\dfrac{x^4}{4}+C\right)\)

D \(y=(x+1)^3-e^x+C\)

×

For solving equations of the form

\(\dfrac{dy}{dx}=h(x)\,g(y)\),  separate the variables and integrate \(\to\;\displaystyle \int\dfrac{dy}{g(y)}=\int h(x)\,dx\)

In this case 

\(\dfrac{dy}{dx}=e^{x+y}+e^yx^3\)

\(\Rightarrow\;\dfrac{dy}{dx}=e^xe^y+e^yx^3=e^y\left[e^x+x^3\right]\)

\(\Rightarrow\;\dfrac{dy}{e^y}=\left(e^x+x^3\right)dx\)

\(\Rightarrow\;e^{-y}dy=(e^x+x^3)\,dx\)

Integrate both sides 

\(\displaystyle\int e^{-y}dy=\displaystyle\int \left(e^x+x^3\right)\,dx\)

\(\Rightarrow\;-e^{-y}=e^x+\dfrac{x^4}{4}+C\)

\(\Rightarrow\;e^{-y}=-e^x-\dfrac{x^4}{4}+C\)

\(\Rightarrow\;y=-ln\left(-e^x-\dfrac{x^4}{4}+C\right)\)

Solve the differential equation \(\dfrac{dy}{dx}=e^{x+y}+e^yx^3\) and \(y\) in terms of \(x\) in the solution.

A

\(y=ln\left(e^x+x^2-x\right)+C\)

.

B

\(y=2sin^2x-cos\,x+C\)

C

\(y=-ln\left(-e^x-\dfrac{x^4}{4}+C\right)\)

D

\(y=(x+1)^3-e^x+C\)

Option C is Correct

Illustration Questions

Solve the initial value problem \(\dfrac{dy}{dx}=2e^{2x}y^2,\;y(0)=-1\)

A \(y=-e^{-2x}\)

B \(y=x\,ln\,x+e^x\)

C \(y=e^{2x}+sin\,x\)

D \(y=x+\dfrac{1}{x}+e^x\)

×

For solving equations of the form

\(\dfrac{dy}{dx}=h(x)\,g(y)\), separate the variables and integrate \(\to\;\displaystyle \int\dfrac{dy}{g(y)}=\int h(x)\,dx\)

In this case 

\(\dfrac{dy}{dx}=2e^{2x}y^2\)

\(\Rightarrow\;\dfrac{dy}{y^2}=2e^{2x}dx\)

 

 

Integrate on both sides 

\(\displaystyle\int \dfrac{dy}{y^2}=\displaystyle\int 2e^{2x}dx\)

\(\Rightarrow\;\displaystyle\int y^{-2}dy=\displaystyle\int 2e^{2x}dx\)

\(\Rightarrow\;-\dfrac{1}{y}=e^{2x}+C\;\to\) General solution

 

Now use initial condition \(\to\;y(0)=-1\)

i.e. when \(x=0,\;y=-1\)

\(\Rightarrow\;\dfrac{-1}{-1}=e^{2×0}+C\)

\(\Rightarrow\;1=1+C\)

\(\Rightarrow\;C=0\)

\(\therefore\) Particular solution is \(-\dfrac{1}{y}=e^{2x}\)

\(\Rightarrow\;y=-e^{-2x}\)

Solve the initial value problem \(\dfrac{dy}{dx}=2e^{2x}y^2,\;y(0)=-1\)

A

\(y=-e^{-2x}\)

.

B

\(y=x\,ln\,x+e^x\)

C

\(y=e^{2x}+sin\,x\)

D

\(y=x+\dfrac{1}{x}+e^x\)

Option A is Correct

Finding Equation of a Curve whose Slope of Tangent at a General Point (x,y) is given

  • Suppose slope of any curve is given in terms of \(x,\,y\), then we equate it to \(f'(x)\) and solve the differential equation obtained. The solution is a family of curves as it contains  C.
  • If the curve passes through a particular point \((x_0,\,y_0)\), C can be found and we get a particular member of the family.

Slope \(\to\) given \(f(x)\) to be found.

Illustration Questions

Find the equation of a family of curve whose slope at the point \((x,\,y)\) is \(\dfrac{3y}{2x}\).

A \(y^2=Cx^3\)

B \(y=C\,ln\,x\)

C \(y^3=Cx^2\)

D \(y=Cx+\dfrac{1}{x}\)

×

Slope at \((x,\,y)=\dfrac{dy}{dx}=f'(x)\)

\(\therefore\) differential equation of the curve is 

\(\dfrac{dy}{dx}=\dfrac{3y}{2x}\)

Which is of variable separable form.

\(\Rightarrow\;\dfrac{1}{y}dy=\dfrac{3}{2x}dx\)

Integrate both sides

\(\displaystyle\int \dfrac{1}{y}dy=\int\dfrac{3}{2x}dx\)

\(\Rightarrow\;ln\,y=\dfrac{3}{2}ln\,x+ln\;C\)

\(\Rightarrow\;ln\,y=\underbrace{ln\left(x^{3/2}C\right)}_\text{properties of log}\)

\(\Rightarrow\;y=C\,x^{3/2}\)

\(\Rightarrow\;y^2=C\,x^3\)

Find the equation of a family of curve whose slope at the point \((x,\,y)\) is \(\dfrac{3y}{2x}\).

A

\(y^2=Cx^3\)

.

B

\(y=C\,ln\,x\)

C

\(y^3=Cx^2\)

D

\(y=Cx+\dfrac{1}{x}\)

Option A is Correct

Finding Particular Member of Family of Curve when it Passes through a Particular Point

  • Suppose slope of any curve is given in terms of \(x,\,y\), then we equate it to \(f'(x)\) and solve the differential equation obtained. The solution is a family of curves as it contains C.
  • If the curve passes through a particular point \((x_0,\,y_0)\), C can be found and we get a particular member of the family.

Slope \(\to\) given \(f(x)\) to be found.

When we solve a first order differential equation, we get a relation between \(y\) and \(x\) and there is an arbitrary constant C in the relation obtained (due to constant of integration).

This means that assigning different values of C will give different curves of similar nature. This is called family of curves.

Out of these curves, there will be only one curve which passes through a particular point, that is called the Particular Member.

Illustration Questions

Find the equation of a curve whose slope of tangent at \(p(x,\,y)\) is \(x(3x-1)\) and which passes through the point \((1,\,1)\).

A \(y=x^3-\dfrac{x^2}{2}+\dfrac{1}{2}\)

B \(y=x\,ln\,x\)

C \(y=x^2-x+1\)

D \(y=sin^2x-x\)

×

Slope at \((x,\,y)=\dfrac{dy}{dx}=f'(x)\)

\(\therefore\) differential equation of the curve is 

\(\dfrac{dy}{dx}=x(3x-1)\)

\(\Rightarrow\;\dfrac{dy}{dx}=3x^2-x\)

which is variable separable form

\(\Rightarrow\;dy=(3x^2-x)\,dx\)

Integrate both sides

\(\displaystyle\int dy=\int \left(3x^2-x\right)\,dx\)

\(\Rightarrow\;y=\dfrac{3x^3}{3}-\dfrac{x^2}{2}+C\rightarrow\)   (General solution which represents family of curve)

It passes through \((1,\,1)\)

 where \(x=1,\;y=1\)

\(\Rightarrow\;1=1-\dfrac{1}{2}+C\)

\(\Rightarrow\;C=\dfrac{1}{2}\)

\(\therefore\) Required curve is  \(y=x^3-\dfrac{x^2}{2}+\dfrac{1}{2}\) 

Find the equation of a curve whose slope of tangent at \(p(x,\,y)\) is \(x(3x-1)\) and which passes through the point \((1,\,1)\).

A

\(y=x^3-\dfrac{x^2}{2}+\dfrac{1}{2}\)

.

B

\(y=x\,ln\,x\)

C

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

D

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

Option A is Correct

Differential Equations in Non Variable Separable Form

  • Consider the differential equation

\(\dfrac{dy}{dx}=ax+by+{c}\to{(1)}\)

  •  This equation is not in variable separable form

 but can be converted into variable separable form by making a substitution

\(ax+by+c=u\;\to\) Differentiate both sides with respect to \(x\)

\(\Rightarrow\;a+\dfrac{b\,dy}{dx}=\dfrac{du}{dx}\)

\(\Rightarrow\;\dfrac{dy}{dx}=\dfrac{\dfrac{du}{dx}-a}{b}\)

  • Now differential equation (1) becomes

\(\dfrac{\dfrac{du}{dx}-a}{b}=u\)

\(\Rightarrow\;\dfrac{du}{dx}=bu+a\;\to\) which is of variable separable form

\(\Rightarrow\,\displaystyle \int \dfrac{1}{bu+a}du=\int dx\)

  • Now integrate to get the general solution

Illustration Questions

Solve the differential equation \(\dfrac{dy}{dx}=2x+y+1\)

A \(ln\,|2x+y+3|=x+C\)

B \(y\,sin\,x=2\,ln\,x+C\)

C \(ln(xy)=x+3+C\)

D \(ln\,|2x+y+1|=x^2+C\)

×

To solve equation of the form

\(\dfrac{dy}{dx}=ax+by+c\)

 put  \(ax+by+c=u\)

In this case \(\dfrac{dy}{dx}=2x+y+1\;\to\) (1)

put \(2x+y+1=u\)

(Differentiate both sides with respect to \(x\))

\(\Rightarrow\;2+\dfrac{dy}{dx}=\dfrac{du}{dx}\)

\(\Rightarrow\;\dfrac{dy}{dx}=\dfrac{du}{dx}-2\)

\(\therefore\) Equation (1) becomes

\(\dfrac{du}{dx}-2=u\)

\(\Rightarrow\;\dfrac{du}{dx}=u+2\)

which is of variable separable form

\(\Rightarrow\;\dfrac{du}{u+2}=dx\)

Integrate both sides

\(\Rightarrow\;\displaystyle \int \dfrac{du}{u+2}=\int dx\)

\(\Rightarrow\;ln\,|u+2|=x+C\)

\(\Rightarrow\;\underbrace{ln\,|2x+y+1+2|}_{u=2x+y+1}=x+C\)

\(\Rightarrow\;ln\,|2x+y+3|=x+C\)

Solve the differential equation \(\dfrac{dy}{dx}=2x+y+1\)

A

\(ln\,|2x+y+3|=x+C\)

.

B

\(y\,sin\,x=2\,ln\,x+C\)

C

\(ln(xy)=x+3+C\)

D

\(ln\,|2x+y+1|=x^2+C\)

Option A is Correct

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