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
    3.0 Differential equation of first order and higher degrees
    4.0 Orthogonal trajectory

2.2 Equations reducible to variable separable 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$$$ \sin (x + y) = x\cos (x + y) + c\cos (x + y) + 1 $$
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