     Question Answers

# 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  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.

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
$\dfrac{{{d}^{2}}y}{d{{x}^{2}}}-3\dfrac{dy}{dx}-5\left[ \dfrac{dy}{dx}-3y \right]=0$
$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}-8\dfrac{dy}{dx}+15y=0$

View Notes
Solution of Differential Equation  Order and Degree of Differential Equations  Exact Differential Equation  Homogeneous Differential Equation  Formation of Differential Equations  Differential Equations For Class 12  First Order Differential Equation  Second-Order Differential Equation  Differential Equations  Differential Equation And Its Types  