Maths > Differential Equations > 2.0 Methods to find the solution of first order and first degree differential equation
Differential Equations
1.0 Introduction
2.0 Methods to find the solution of first order and first degree 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
3.0 Differential equation of first order and higher degrees
4.0 Orthogonal trajectory
2.2 Equations reducible to variable separable form
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
In some differential equations, it is not easy to separate the variables but by some substitution, equations can be reduced into a variable separable form.
The general form of this type of differential equations is $$\frac{{dy}}{{dx}} = f(ax + by + c)\quad (a,b \ne 0)$$ which can be solved by putting $ax + by + c = t$.
Question 6. Solve $${\sin ^{ - 1}}\left( {\frac{{dy}}{{dx}}} \right) = x + y$$
Solution:
$$ \frac{{dy}}{{dx}} = \sin (x + y)\quad ...(1) $$ $$ {\text{Put }}x + y = t$$and differentiate it w.r.t $x$ we get $$ {\text{1 + }}\frac{{dy}}{{dx}} = \frac{{dt}}{{dx}}$$Put the value form $(1)$ $$ \frac{{dt}}{{dx}} = \sin t + 1 $$ $$ \frac{{dt}}{{\sin t + 1}} = dx$$Integrating both sides$$ \int {\frac{{dt}}{{\sin t + 1}}} = \int {dx} $$ $$ \int {\frac{{1 - \sin t}}{{{{\cos }^2}t}}} dt = x + c $$ $$ \int {({{\sec }^2}t} - \sec t\tan t)dt = x + c $$ $$ \tan t - \sec t = x + c $$ $$ - \frac{{1 - \sin t}}{{\cos t}} = x + c $$ $$ \sin t - 1 = x\cos t + c\cos t$$substitute the value of$t$$$ \sin (x + y) = x\cos (x + y) + c\cos (x + y) + 1 $$