Question

In each of the Exercises $1$ to $5,$ form a differential equation representing the given family of curves by eliminating arbitrary constants $a$ and $b.$
(1) $\frac{x}{a}+\frac{y}{b}=1$ (2) $y^2=a(b^2-x^2)$ (3) $y=a{e}^{3x}+b{e}^{-2x}$
(4) $y=e^{2x}(a+bx)$ (5) $y=e^x(a\cos x+b\sin x)$

Answer 1

pm$\left(1\right)$ $\frac{x}{a}+\frac{y}{b}=1$

$\frac{1}{a}+\frac{dy}{dx/b}0$

$\frac{dy}{dx}=\frac{-b}{a}$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{2xy}{ab}=1$

$\frac{b^2}{a^2}{x}^2+y^2+\frac{2xyb}{a}=1$

$x^2{\left(\frac{dy}{dx}\right)}^2+y^2-2xy\frac{dy}{dx}=1$

$\left(2\right)$ $y^2=a(b^2-x^2)$

$2y{y}^{\prime }=c-2ax$

$2y{y}^{\prime \prime }+2y^{12}-2a$

$a=y{y}^{\prime \prime }+(y{)}^2$

$2y{y}^{\prime }=-2(y{y}^{\prime \prime }+(y^{\prime }{)}^2)x$

$\left(3\right)$ $y=a{e}^{3x}+b{e}^{-2x}$

$(y^{\prime }=3a{e}^{3x}-2b{e}^{-2x})2$

$y^{\prime \prime }=3a{c}^{3x}+4b{e}^{-2x}$

$2y^{\prime }=6a{e}^{3x}-4b{e}^{-2x}$

$2y^{\prime }+y^{\prime \prime }=15a{e}^{3x}$

$\Rightarrow a=\frac{2y^{\prime }+2y^{\prime \prime }}{15e^{3x}}$

$y^{\prime \prime }=\frac{9}{15}(2y^{\prime }+y^{\prime \prime })+4b{e}^{-2x}$

$3x$ & $5y^{\prime }=6y^{\prime }+3y^{\prime \prime }+4b{e}^{-x}$

$b=\frac{e^{3x}5y^{\prime \prime }-6y^{\prime }+3y^{\prime \prime }}{9b}$

$\left(4\right)$ $y=e^{2x}(a+bx)$

$y=e^{2x}a+b\times e^{2x}$

$y=b{e}^{2x}+b{e}^{2x}+2bx{e}^{2x}$

$y^{\prime }=(2a+2bx)e^{2x}$

$y^{\prime }=2y{e}^{2x}$.

$\left(5\right)$ $y=e^x(a\cos x+b\sin x)$

$y^{\prime }=e^x(a\cos x+b\sin x)+e^x(a\sin x+b\cos x)$

$y^{\prime \prime }=e^x(a\cos x+b\sin x)+e^x(-a\sin x+b\cos x)+e^x(-a\sin x+b\cos x)+e^x(-a\cos x-b\sin x)$

$y^{\prime \prime }=e^x(a\cos x+b\sin x)\pm ,ey$

$y^{\prime \prime }=y-2y^{\prime }$.