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

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 \(x\).

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

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

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

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

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

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

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

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

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

- 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.

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

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

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

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

- 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.

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

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

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

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

- 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

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

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

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

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