2.8 Equations reducible to linear form

1.1 Order and degree of a differential equation
1.2 Formation of a differential equation

2.1 Variable separable method
2.2 Equations reducible to variable separable form
2.3 Homogeneous differential equation
2.4 Equations reducible to homogeneous form
2.5 Exact differential equation
2.6 Equations reducible to exact form
2.7 Linear differential equation
2.8 Equations reducible to linear form

(i) By change of variable: The differential equations which are not in linear form can be converted by suitable substitution of non-linear term.

(ii) Bernoulli's equation: An equation of the form $$\frac{{dy}}{{dx}} + Py = Q{y^n}$$ where $P$ and $Q$ are functions of $x$ only, is known as Bernoulli's equation. In order to solve this equation, it is first converted to a linear form by dividing the equation by ${y^n}$ i.e., $${y^{ - n}}\frac{{dy}}{{dx}} + P{y^{1 - n}} = Q$$

Now, put ${y^{1 - n}} = v \Rightarrow (1 - n){y^{ - n}}\frac{{dy}}{{dx}} = \frac{{dv}}{{dx}}$ i.e., $$\frac{{dv}}{{dx}} + (1 - n)Pv = (1 - n)Q$$ which is a linear equation in $v$. The solution can be obtained using integrating factor as mentioned earlier.

Question 12. Solve $${\sec ^2}y\frac{{dy}}{{dx}} + 2x\tan y = {x^3}$$

Solution: Let us assume $\tan y = z$. Therefore, ${\sec ^2}y.\frac{{dy}}{{dx}} = \frac{{dz}}{{dx}}$. Thus, the given non-linear equation is converted to linear differential equation i.e., $$\frac{{dz}}{{dx}} + 2x.z = {x^3}$$ $$\therefore {\text{I}}{\text{.F}}{\text{. = }}{e^{\int {2xdx} }} = {e^{{x^2}}}$$

The solution is $$z.{e^{{x^2}}} = \int {{x^3}.} {e^{{x^2}}}dx + C$$ Put the value of $z$, we get $$\tan y.{e^{{x^2}}} = \frac{1}{2}\int {{x^2}.} {e^{{x^2}}}.(2x)dx + C$$ Put $t = {x^2}$, we get $$\begin{equation} \begin{aligned} \tan y.{e^{{x^2}}} = \frac{1}{2}\int {t.} {e^t}dt + C = \frac{1}{2}\left( {t.{e^t} - {e^t}} \right) + C \\ \tan y = C{e^{ - {x^2}}} + \frac{{{e^{ - {x^2}}}}}{2}.{e^{{x^2}}}({x^2} - 1) \\ \tan y = C{e^{ - {x^2}}} + \frac{1}{2}\left( {{x^2} - 1} \right) \\\end{aligned} \end{equation} $$

Question 13. Solve $$\frac{{dy}}{{dx}} - \frac{y}{x} = \frac{{{y^2}}}{{{x^2}}}$$

Solution: Divide both sides by ${y^2}$, we get $$\frac{1}{{{y^2}}}\frac{{dy}}{{dx}} - \frac{1}{{xy}} = \frac{1}{{{x^2}}}$$ Put $\frac{1}{y} = t \Rightarrow - \frac{1}{{{y^2}}}\frac{{dy}}{{dx}} = \frac{{dt}}{{dx}}$

Therefore, the given bernoulli's equation becomes $$ - \frac{{dt}}{{dx}} - \frac{t}{x} = \frac{1}{{{x^2}}}$$ which is a linear differential equation.

Therefore, Integrating factor (I.F.) is $${e^{\int {\frac{1}{x}dx} }} = {e^{\log x}} = x$$

Now, general solution is $$\begin{equation} \begin{aligned} t.x = \int { - \frac{1}{{{x^2}}}.xdx} + C \\ tx = - \ln x + C \\ \frac{x}{y} = - \ln x + C \\\end{aligned} \end{equation} $$