Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
Advertisement Remove all ads

Solve the Following Differential Equation:- ( X + Y ) D Y D X = 1 - Mathematics

Sum

Solve the following differential equation:-

\[\left( x + y \right)\frac{dy}{dx} = 1\]

Advertisement Remove all ads

Solution

We have,

\[\left( x + y \right)\frac{dy}{dx} = 1\]

\[ \Rightarrow \frac{dy}{dx} = \frac{1}{\left( x + y \right)}\]

Let x + y = v

\[ \Rightarrow 1 + \frac{dy}{dx} = \frac{dv}{dx}\]

\[ \Rightarrow \frac{dy}{dx} = \frac{dv}{dx} - 1\]

\[ \therefore \frac{dv}{dx} - 1 = \frac{1}{v}\]

\[ \Rightarrow \frac{dv}{dx} = \frac{1}{v} + 1\]

\[ \Rightarrow \frac{dv}{dx} = \frac{1 + v}{v}\]

\[ \Rightarrow \frac{v}{1 + v}dv = dx\]

Integrating both sides, we get

\[\int\frac{v}{1 + v}dv = \int dx\]

\[ \Rightarrow \int\frac{v + 1 - 1}{1 + v}dv = \int dx\]

\[ \Rightarrow \int dv - \int\frac{1}{1 + v}dv = \int dx\]

\[ \Rightarrow v - \log \left| v + 1 \right| = x - \log C\]

\[ \Rightarrow x + y - \log \left| x + y + 1 \right| = x - \log C\]

\[ \Rightarrow y - \log \left| x + y + 1 \right| = - \log C\]

\[ \Rightarrow y = \log\left| x + y + 1 \right| - \log C\]

\[ \Rightarrow y = \log\left| \frac{x + y + 1}{C} \right|\]

\[ \Rightarrow C e^y = x + y + 1\]

  Is there an error in this question or solution?
Advertisement Remove all ads

APPEARS IN

RD Sharma Class 12 Maths
Chapter 22 Differential Equations
Revision Exercise | Q 66.13 | Page 147
Advertisement Remove all ads
Advertisement Remove all ads
Share
Notifications

View all notifications


      Forgot password?
View in app×