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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 $$x$$.

#### 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

#### 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

#### 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

#### 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.

#### 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.

#### 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

#### 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