Differential Equation flash cards

1. 1.Define order and degree of a differential equation.

   Order = highest order derivative present; Degree = power of the highest order derivative, after the equation is made free from radicals/fractions in derivatives (must be a polynomial in derivatives).

   Hint: Degree is defined only if the DE is polynomial in derivatives.

2. 2.Solve a homogeneous differential equation $\frac{dy}{dx}=f\left(\frac{y}{x}\right)$.

   Substitute $y=vx \Rightarrow \frac{dy}{dx}=v+x\frac{dv}{dx}$, giving $x\frac{dv}{dx}=f(v)-v$, a variable-separable equation in $v,x$.

   Hint: Use $v=y/x$ substitution.

3. 3.State the solution (integrating factor method) of a linear differential equation $\frac{dy}{dx}+Py=Q$, where $P,Q$ are functions of $x$.

   $I.F.=e^{\int P\,dx}$ and solution is $y\cdot(I.F.)=\int Q\cdot(I.F.)\,dx+C$.

   Hint: Multiply throughout by $e^{\int P dx}$.

4. 4.State the solution of the linear differential equation $\frac{dx}{dy}+P_1x=Q_1$, where $P_1,Q_1$ are functions of $y$.

   $I.F.=e^{\int P_1\,dy}$ and solution is $x\cdot(I.F.)=\int Q_1\cdot(I.F.)\,dy+C$.

   Hint: Treat $x$ as dependent variable, $y$ as independent.

5. 5.How is the differential equation of a family of curves with $n$ arbitrary constants obtained, and what does its order equal?

   Differentiate the family's equation $n$ times and eliminate the $n$ arbitrary constants; the resulting DE has order $=n$.

   Hint: Number of constants eliminated = order of DE.

6. 6.Form the differential equation of the family $y=Ae^{3x}+Be^{-3x}$ by eliminating the arbitrary constants $A,B$.

   Differentiating twice: $y'=3Ae^{3x}-3Be^{-3x}$ and $y''=9Ae^{3x}+9Be^{-3x}=9(Ae^{3x}+Be^{-3x})=9y$. So the required DE is $y''-9y=0$.

   Hint: Differentiate twice and notice $y''$ reproduces $9y$ directly.

7. 7.Form the differential equation of the family of circles touching the x-axis at the origin: $x^2+(y-a)^2=a^2$.

   Expand to $x^2+y^2=2ay$, so $a=\dfrac{x^2+y^2}{2y}$. Differentiating the original: $2x+2yy'=2ay'$, i.e. $x+yy'=ay'$. Substituting $a$ and simplifying gives $y'(x^2-y^2)=2xy$, i.e. $(x^2-y^2)\\dfrac{dy}{dx}=2xy$.

   Hint: Eliminate $a$ using both the equation and its derivative.

8. 8.Form the differential equation representing the family of parabolas $y^2=4a(x+a)$ (eliminate the single parameter $a$).

   Differentiating: $2yy'=4a\Rightarrow a=\dfrac{yy'}{2}$. Substituting into $y^2=4ax+4a^2$ gives $y^2=2xyy'+y^2(y')^2$; dividing by $y$ yields the DE $y(y')^2+2xy'-y=0$.

   Hint: Solve the original equation for $a$ from its derivative, then resubstitute.

9. 9.Form the differential equation of $y=c_1\cos2x+c_2\sin2x$ by eliminating $c_1,c_2$.

   $y'=-2c_1\sin2x+2c_2\cos2x$ and $y''=-4c_1\cos2x-4c_2\sin2x=-4y$. Hence the DE is $y''+4y=0$.

   Hint: Two constants need two differentiations; look for the equation reproducing $y$ itself.

10. 10.Eliminate the arbitrary constants $a,b$ from $y=ax^2+bx$ to form the differential equation.

    $y'=2ax+b$, $y''=2a\Rightarrow a=\dfrac{y''}{2}$, and $b=y'-2ax=y'-xy''$. Substituting into $y=ax^2+bx$: $y=\dfrac{x^2}{2}y''+x(y'-xy'')=xy'-\dfrac{x^2}{2}y''$. Multiplying by 2 gives $x^2y''-2xy'+2y=0$.

    Hint: Find $a$ from $y''$, then $b$ from $y'$, and resubstitute into $y$.

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