Question

The differential equation of the family of curves $A{{e}^{3x}}+B{{e}^{5x}}$ where A and B are arbitrary constants is:
[a] $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+8\dfrac{dy}{dx}+15y=0$
[b] $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-\dfrac{dy}{dx}+y=0$
[c] $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-8\dfrac{dy}{dx}+15y=0$
[d] None of the above

Answer 1

Hint: Try removing the variables A and B by differentiating twice and solving for A and B. Alternatively you can remove by finding the value of A in terms of B, y and x. Then differentiate once and find the value of B in terms of x, y, $\dfrac{dy}{dx}$ and Differentiate again.

Complete step-by-step answer:
Let $y=A{{e}^{3x}}+B{{e}^{5x}}$
Differentiating both sides, we get
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( A{{e}^{3x}}+B{{e}^{5x}} \right)$
We know that $\dfrac{d}{dx}\left( f(x)+g(x) \right)=\dfrac{d}{dx}\left( f(x) \right)+\dfrac{d}{dx}\left( g(x) \right)$
Using the above property, we get
$\dfrac{dy}{dx}=\dfrac{d}{dx}\left( A{{e}^{3x}} \right)+\dfrac{d}{dx}\left( B{{e}^{5x}} \right)$
We know that $\dfrac{d}{dx}\left( Cg(x) \right)=C\dfrac{d}{dx}g(x)$
Using we get
$\dfrac{dy}{dx}=A\dfrac{d}{dx}{{e}^{3x}}+B\dfrac{d}{dx}{{e}^{5x}}$
Chain rule of differentiation $\dfrac{d}{dx}f\left( g\left( x \right) \right)=\dfrac{d}{d\left( g(x) \right)}f\left( g\left( x \right) \right)\times \dfrac{d}{dx}g(x)$
Using chain rule, we get
$\dfrac{dy}{dx}=A\dfrac{d}{d(3x)}{{e}^{3x}}\dfrac{d}{dx}(3x)+B\dfrac{d}{d(5x)}{{e}^{5x}}\dfrac{d}{dx}(5x)$
We know that $\dfrac{d}{dt}{{e}^{t}}={{e}^{t}}$
\[\Rightarrow \dfrac{dy}{dx}=A{{e}^{3x}}\dfrac{d}{dx}3x+B{{e}^{5x}}\dfrac{d}{dx}5x\]
$\Rightarrow \dfrac{dy}{dx}=A{{e}^{3x}}\times 3+B{{e}^{5x}}\times 5$
$\Rightarrow \dfrac{dy}{dx}=3A{{e}^{3x}}+5B{{e}^{5x}}\text{ (i)}$
Differentiating again we get
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( 3A{{e}^{3x}}+5B{{e}^{3x}} \right)$
We know that $\dfrac{d}{dx}\left( f(x)+g(x) \right)=\dfrac{d}{dx}\left( f(x) \right)+\dfrac{d}{dx}\left( g(x) \right)$
Using the above property, we get
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\dfrac{d}{dx}\left( 3A{{e}^{3x}} \right)+\dfrac{d}{dx}\left( 5B{{e}^{5x}} \right)$
We know that $\dfrac{d}{dx}\left( Cg(x) \right)=C\dfrac{d}{dx}g(x)$
Using we get
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=3A\dfrac{d}{dx}{{e}^{3x}}+5B\dfrac{d}{dx}{{e}^{5x}}\]
Using chain rule, we get
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=3A\dfrac{d}{d(3x)}{{e}^{3x}}\dfrac{d}{dx}3x+5B\dfrac{d}{d(5x)}{{e}^{5x}}\dfrac{d}{dx}5x\]
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=3A{{e}^{3x}}\times 3+5B{{e}^{5x}}\times 5$
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=9A{{e}^{3x}}+25B{{e}^{5x}}\text{ (ii)}$
Multiplying equation (i) by 3 and subtracting (i) from (ii) we get
\[\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-3\dfrac{dy}{dx}=9A{{e}^{3x}}-9A{{e}^{3x}}+25B{{e}^{5x}}-15B{{e}^{5x}}\]
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}-3\dfrac{dy}{dx}=10B{{e}^{5x}}\]
Dividing both sides by $10{{e}^{5x}}$
$\Rightarrow \dfrac{1}{10}\left[ \dfrac{{{d}^{2}}y}{d{{x}^{2}}}-3\dfrac{dy}{dx} \right]{{e}^{-5x}}=B\text{ (iii)}$
Multiplying equation (i) by 5 and subtracting (i) from (ii) we get
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-5\dfrac{dy}{dx}=9{{A}^{3x}}-15{{A}^{3x}}+25B{{e}^{3x}}-25B{{e}^{3x}}$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}-5\dfrac{dy}{dx}=-6A{{e}^{3x}}$
Dividing both sides by $-6{{e}^{3x}}$ we get
$\dfrac{-1}{6}\left[ \dfrac{{{d}^{2}}y}{d{{x}^{2}}}-5\dfrac{dy}{dx} \right]{{e}^{-3x}}=A\text{ (iv)}$
Put the value of B and A from (iii) and (iv) in the family equation we get
$y=\dfrac{-1}{6}\left[ \dfrac{{{d}^{2}}y}{d{{x}^{2}}}-5\dfrac{dy}{dx} \right]{{e}^{-3x}}{{e}^{3x}}+\dfrac{1}{10}\left[ \dfrac{{{d}^{2}}y}{d{{x}^{2}}}-3\dfrac{dy}{dx} \right]{{e}^{-5x}}{{e}^{5x}}$
$\Rightarrow y=\dfrac{-1}{6}\left[ \dfrac{{{d}^{2}}y}{d{{x}^{2}}}-5\dfrac{dy}{dx} \right]+\dfrac{1}{10}\left[ \dfrac{{{d}^{2}}y}{d{{x}^{2}}}-3\dfrac{dy}{dx} \right]$
Multiplying both sides by 30 we get
$30y=-5\left[ \dfrac{{{d}^{2}}y}{d{{x}^{2}}}-5\dfrac{dy}{dx} \right]+3\left[ \dfrac{{{d}^{2}}y}{d{{x}^{2}}}-3\dfrac{dy}{dx} \right]$
Using distributive law $a(b+c)=ab+ac$
$\Rightarrow 30y=-5\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+25\dfrac{dy}{dx}+3\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-9\dfrac{dy}{dx}$
$\Rightarrow 30y=-2\dfrac{{{d}^{2}}y}{d{{x}^{2}}}+16\dfrac{dy}{dx}$
Dividing both sides by 2 and transposing all terms to LHS we get
$15y+\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-8\dfrac{dy}{dx}=0$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}-8\dfrac{dy}{dx}+15y=0$
Option (c) is correct.

Note: This question can also be solved by using the following property
If $y=A{{e}^{mx}}+B{{e}^{nx}}$ then the differential equation of the family of the curves is $f(D)=0$ where
$f(x)$ is the quadratic with roots m and n.
Here m = 3 and n = 5.
So, f(x) = (x-3)(x-5) = ${{x}^{2}}-8x+15$
Hence the differential equation of the family of curves is f(D) = ${{D}^{2}}-8D+15=0$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}-8\dfrac{dy}{dx}+15=0$
Alternatively, we have $y=A{{e}^{3x}}+B{{e}^{5x}}$
$\Rightarrow y-B{{e}^{5x}}=A{{e}^{3x}}$
\[\Rightarrow \left( y-B{{e}^{5x}} \right){{e}^{-3x}}=A\]
Differentiating both sides w.r.t x once we get
$\dfrac{d}{dx}\left( \left( y-B{{e}^{5x}} \right){{e}^{-3x}} \right)=\dfrac{dA}{dx}$
Using product rule of differentiation $\dfrac{d}{dx}uv=v\dfrac{du}{dx}+u\dfrac{dv}{dx}$ we get
${{e}^{-3x}}\dfrac{d\left( y-B{{e}^{5x}} \right)}{dx}+\left( y-B{{e}^{5x}} \right)\dfrac{d}{dx}{{e}^{-3x}}=0$
$\Rightarrow {{e}^{-3x}}\left[ \dfrac{dy}{dx}-5B{{e}^{5x}} \right]-3\left( y-B{{e}^{5x}} \right){{e}^{-3x}}=0$
Taking ${{e}^{-3x}}$ common and dividing both sides by ${{e}^{-3x}}$
$\dfrac{{{e}^{-3x}}}{{{e}^{-3x}}}\left[ \dfrac{dy}{dx}-5B{{e}^{5x}}-3y+3B{{e}^{5x}} \right]=\dfrac{0}{{{e}^{-3x}}}$
$\dfrac{dy}{dx}-2B{{e}^{5x}}-3y=0$
$\Rightarrow \dfrac{dy}{dx}-3y=2B{{e}^{5x}}$
$\Rightarrow \dfrac{1}{2}\left[ \dfrac{dy}{dx}-3y \right]{{e}^{-5x}}=B$
Differentiating again we get
$\dfrac{1}{2}\dfrac{d}{dx}\left[ \left( \dfrac{dy}{dx}-3y \right){{e}^{-5x}} \right]=0$
Using product rule of differentiation
$\dfrac{1}{2}{{e}^{-5x}}\dfrac{d}{dx}\left( \dfrac{dy}{dx}-3y \right)+\dfrac{1}{2}\left[ \dfrac{dy}{dx}-3y \right]\dfrac{d}{dx}{{e}^{-5x}}=0$
Taking $\dfrac{{{e}^{-5x}}}{2}$ and transposing to RHS we get
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-3\dfrac{dy}{dx}-5\left[ \dfrac{dy}{dx}-3y \right]=0$
\[\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}-8\dfrac{dy}{dx}+15y=0\]