Learn definition of a Differential Equation & order of differential equations, Practice to find particular solution of the differential equation problems.
e.g (1) \(\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}=sin\,x\)
(2) \(\dfrac{2ydy}{dx}+lnx=sin^2\,x\)
(3) \({\sqrt{1+\dfrac {dy}{dx}}}=\dfrac{d^2y}{dx^2}\) are all differential equations.
Example: (1) \(\dfrac{d^2y}{dx^2}+\dfrac{xdy}{dx}=5sin\,x\) is of order 2. \(\left(\because\,\dfrac{d^2y}{dx^2}\text{represents order 2}\right)\)
(2) \(x^3\dfrac{dy}{dx}+lnx=5\) is of order 1. \(\left(\because\,\dfrac{dy}{dx}\text{represents order 1}\right)\)
(3) \(5y^3\dfrac{d^3y}{dx^3}=2x+\dfrac{dy}{dx}=5sin\,x\) is of order 3. \(\left(\because\,\dfrac{d^3y}{dx^3}\text{represents order 3}\right)\)
A \(2y^3–x^2=sin\,x\)
B \(cos^3x+e^4\,lnx=2\)
C \(5y+x=7\)
D \(\dfrac{xdy}{dx}+y=12cos^3x\)
A \(k=\pm\dfrac{5}{4}\)
B \(k=\pm{3}\)
C \(k=\pm{9}\)
D \(k=\pm\dfrac{7}{6}\)
When the solution to a differential equation is asked usually all solutions have to be reported
\(f'(x)=f(x)\to\) we know that the solution to this equation is \(f(x)=e^x\) but all function of the form \(f(x)=e^x\) are also have solutions we say that \(y=ce^x\) is called the general solution (or family of solution) to the differential equation.
e.g. \(y'=y\to y(0)=1^{ \nearrow^{\text {initial condition}}}\)
\(\Rightarrow\,y=ce^x \to\) general solution
Use initial condition
\(1=ce^0\Rightarrow\,c=1\)
\(\therefore\,y=e^x \to\) particular solution.
A \(y=2sinx+lnx\)
B \(y=e^x(x+1)\)
C \(y=ce^x+4\) (C \(\in\)R)
D \(y=e^x(x^2+1)\)
\(f(x)\) is a solution to \(y'=5y\) if
\(f'(x)=5f(x)\)
A \(f(x)=y=sin^2x\)
B \(f(x)=y=cos^2x\)
C \(f(x)=y=2e^{3x}\)
D \(f(x)=y=ln2x\)
Suppose we desire to know the value of \(\alpha\) for which \(y=e^{\alpha x}\) is a solution to the differential equation.
\(y''+2y'+y=0\)
Find \(y'=\alpha e^{\alpha x}\) and \(y''=\alpha ^2e^{\alpha x}\) and put in the equation.
\(\alpha^2 e^{\alpha x}+2\alpha e^{\alpha x}+e^{\alpha x}=0\)
\(\Rightarrow \,e^{\alpha x}(\alpha^2+2\alpha+1)=0\)
\(\Rightarrow\,e^{\alpha x}(\alpha+1)^2=0\)
\(\Rightarrow\,\alpha=–1\)
A \(\alpha=-2,\,\alpha=-3\)
B \(\alpha=-5,\,\alpha=\dfrac{1}{2}\)
C \(\alpha=1,\,\alpha=6\)
D \(\alpha=4,\,\alpha=1\)