Answer to Question #295777 in Differential Equations for Momi

Question

Solve the following first order linear equations

1)dy/dx=e^2x+3y

2)xy×dy/dx=2(y+3)

3)(1+x^2).dy/dx-xy(1+y)=0

4)sinx/1+y.dy/dx=cosx

5)cos^2x.dy/dx=y-3

Answer 1

1)

$\displaystyle \frac{dy}{dx}-3y=e^{2x}$

Integrating factor $\displaystyle =e^{-3x}$

Multiplying both sides of the given DE yields;

$\displaystyle \frac{d}{dx}(ye^{-3x})=e^{-x}\\ ye^{-3x}=\int e^{-x}\ dx=-e^{-x}+c,$ where c is an arbitrary constant.

$\displaystyle \Rightarrow y=ce^{3x}-e^{2x}$

2)

$\displaystyle xy\frac{dy}{dx}=2(y+3)$

By method of separation of variables,

$\displaystyle \frac{y}{y+3}\frac{dy}{dx}=\frac{2}{x}$

Integrating both sides wrt x yields;

$\displaystyle \int\frac{y}{y+3}\frac{dy}{dx}\ dx=\int\frac{2}{x}\ dx\\ \Rightarrow \int\frac{y}{y+3}\ dy=\int\frac{2}{x}\ dx\\ \Rightarrow\int\left(1-\frac{3}{y+3}\right)\ dy=\int\frac{2}{x}\ dx\\ \Rightarrow y-3\ln(y+3)=2\ln(x)+\ln(c)$, where c is an arbitrary constant

$\displaystyle \Rightarrow y=cx^2(y+3)^3$

3)

$\displaystyle (1+x^2)\frac{dy}{dx}-xy(1+y)=0\\ \text{By method of separation of variables,}\\ \Rightarrow \frac{1}{y(y+1)}\frac{dy}{dx}=\frac{x}{1+x^2}\\ \Rightarrow \left(\frac{1}{y}-\frac{1}{y+1}\right)\frac{dy}{dx}=\frac{x}{1+x^2}$, by partial fraction

integrating both sides wrt x yields;

$\displaystyle \int\left(\frac{1}{y}-\frac{1}{y+1}\right)\frac{dy}{dx}\ dx=\int\frac{x}{1+x^2}\ dx\\ \Rightarrow\int\frac{1}{y}\ dy-\int\frac{1}{y+1}\ dy=\int\frac{x}{1+x^2}\ dx\\ \Rightarrow \ln(y)-\ln(y+1)=\frac{1}{2}\ln(x^2+1)+\ln(c)\\$, where c is an arbitrary constant

$\displaystyle \frac{y}{y+1}=c\sqrt{x^2+1}$

4)

$\displaystyle \frac{\sin x}{1+y}\cdot\frac{dy}{dx}=\cos x\\$

By method of separation of variables,

$\displaystyle \Rightarrow \frac{1}{1+y}\frac{dy}{dx}=\frac{\cos x}{\sin x}$

integrating both sides wrt x yields;

$\displaystyle \int\frac{1}{1+y}\frac{dy}{dx}\ dx=\int\frac{\cos x}{\sin x}\ dx\\ \Rightarrow \int\frac{1}{1+y}\ dy=\int\frac{\cos x}{\sin x}\ dx\\ \Rightarrow\ln(y+1)=\ln(\sin x)+\ln(c)$, where c is an arbitrary constant

$\displaystyle \Rightarrow y=c\sin x-1$

5)

$\displaystyle \cos^2(x)\frac{dy}{dx}=y-3$

By method of separation of variables,

$\displaystyle \Rightarrow \frac{1}{y-3}\frac{dy}{dx}=\frac{1}{\cos^2 x}$

integrating both sides wrt x yields;

$\displaystyle \Rightarrow \int\frac{1}{y-3}\frac{dy}{dx}\ dx=\int\frac{1}{\cos^2 x}\ dx\\ \Rightarrow \int\frac{1}{y-3}\ dy=\int\frac{1}{\cos^2 x}\ dx\\ \Rightarrow \int\frac{1}{y-3}\ dy=\int\sec^2 x\ dx\\ \Rightarrow \ln(y-3)=\tan x+c,$where c is an arbitrary constant.

$\displaystyle \Rightarrow y=3+Ae^{\tan (x)}$, where $A=e^c$