Differential Equations Test

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Differential Equations Test - 53

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Weekly Quiz Competition

Question 1
1 / -0

If $$y_1(x)$$ is a solution of the differential equation $$\frac{dy}{dx} + f(x) + y = 0$$, then a solution of differential equation $$\frac{dy}{dx} + f(x) + y = r (x)$$ is

A
$$\frac{1}{y(x)} \int y_1(x) dx$$

B
$$y_1(x) \int \frac{r(x)}{y_1(x)} dx$$

C
$$ \int r(x)y_1(x) dx$$

D
none of these

Solution
Question 2
1 / -0

Solution of differential equation $$\frac{dy}{dx} = sin(x+y) + cos(x+y)$$ is

A
$$log|1+tan \frac{x+y}{2}| = y+c$$

B
$$log|2+sec \frac{x+y}{2}| = x+c$$

C
$$log|1+tan (x+y)| = y+c$$

D
none of these

Solution
Question 3
1 / -0

Solution of differential equation $$\dfrac{dy}{dx} = \sin (x+y) + \cos(x+y)$$ is

A
$$log |1 + \tan\frac{x+y}{2} | = y+c$$

B
$$log |1 + \sec\frac{x+y}{2} | = x+c$$

C
$$log |1 + \tan(x+y) | = y+c$$

D
none of these

Solution
Question 4
1 / -0

The general solution of $$\dfrac {dy}{dx} = \dfrac {ax + h}{by + k}$$ represents a circle only when

A
$$a = b = 0$$

B
$$a = -b\neq 0$$

C
$$a = b \neq 0, h = k$$

D
$$a = b \neq 0$$

Solution

Given that
$$ \dfrac{dy}{dx} = \dfrac{ax + h}{by + k}$$

$$ \Rightarrow dy \; (by + k) = dx \; (ax + h)$$

Integrating LHS and RHS,

$$ b\dfrac{y^2}{2} + ky + c = a\dfrac{x^2}{2} + h x + c $$

$$\Rightarrow ax^2 - b y^2 + 2hx - 2ky + c = 0 $$

The general equation of a cirlce is given as:
$$ x^2 + y^2 + 2gx + 2fy + c = 0$$

Upon comparing the obtained equation with the general equation of a circle,
$$ ax^2 - b y^2 + 2hx - 2ky + c = 0 $$
represents a cirlce if
$$ a = -b \neq 0 $$.
Question 5
1 / -0

$$Solution\quad of\quad the\quad equation:\quad (1-{ x }^{ 2 })dy\quad +\quad xydx=\quad x{ y }^{ 2 }dx$$

A
$$({ y-1 }^{ )2 }{ (1- }{ x }^{ 2 })=0$$

B
$$({ y-1 }^{ )2 }{ (1- }{ x }^{ 2 })={ c }^{ 2 }{ y }^{ 2 }$$

C
$$({ y-1 }^{ )2 }{ (1+ }{ x }^{ 2 })={ c }^{ 2 }{ y }^{ 2 }$$

D
None of these

Solution

$$(1-x^{2})dy+xydx=xy^{2}dx$$
Dividing by $$y^{2}dx$$, we get(and also divisible by $$(1-x^{2})$$
$$\cfrac{1}{y^{2}}\cfrac{dy}{dx}+\cfrac{x}{1-x^{2}}\cfrac{1}{y}=\cfrac{x}{1-x^{2}}$$
Now by putting $$\cfrac{-1}{y}=z$$
$$\cfrac{1}{y^{2}}dy=dz$$
Therefore, equation becomes
$$\cfrac{dz}{dx}-\cfrac{x}{1-x^{2}}dz=\cfrac{x}{(1-x^{2})}$$
Now ,$$IF=e^{\int\cfrac{-x}{1-x^{2}}}=e^{+\cfrac{1}{2}\log(1-x^{2})}=\sqrt{1-x^{2}}$$
Multiplying by IF , we get
$$\cfrac{d}{dx}((\sqrt{1-x^{2}})z)$$=$$\cfrac{x}{\sqrt{1-x^{2}}}$$
Now integrating on both the sides and taking
$$\sqrt{1-x^{2}}=P$$ on RHS, we get
$$\cfrac{1}{2\sqrt{1-x^{2}}}\times(-2x)dx=dP$$
$$\cfrac{-x}{\sqrt{1-x^{2}}}dx=dP$$
$$\cfrac{x}{\sqrt{1-x^{2}}}dx=-dP$$
$$\int d((1-\sqrt{1-x^{2}})dz)=-\int-dP$$
$$(1-\sqrt{1-x^{2}})z=-P+C=-\sqrt{1-x^{2}}+C$$
$$(\sqrt{1-x^{2})}(z+1)=C$$
$$\int(\sqrt{1-x^{2})}(\cfrac{-1}{y}+1)=C$$
Now squaring on both the sides, we get
$$(y-1)^{2}(1-x^{2})=C^{2}y^{2}$$
Question 6
1 / -0

The solution of the differential equation $$x^{2} \dfrac{dy}{dx} = x^{2} + xy + y^{2}$$ is-

A
$$tan^{-1}\dfrac{y}{x} = log x + c$$

B
$$tan^{-1}\dfrac{y}{x} = -log x + c$$

C
$$sin^{-1}\dfrac{y}{x} = log x + c$$

D
$$tan^{-1}\dfrac{x}{y} = log x + c$$

Solution

$$x^{2}\cfrac{dy}{dx}=x^{2}+xy+y^{2}$$
$$\cfrac{dy}{dx}=1+\cfrac{y}{x}+(\cfrac{y}{x})^{2}$$
Let $$\cfrac{y}{x}=v$$
$$y=vx$$
$$\cfrac{dy}{dx}=v+x\cfrac{dv}{dx}$$
$$v+x\cfrac{dv}{dx}=1+v+v^{2}$$
$$x\cfrac{dv}{dx}=1+v^{2}$$
$$\cfrac{dv}{1+v^{2}}=\cfrac{dx}{x}$$
$$\tan^{-1}\,v=\log x+C$$
$$\tan^{-1}\,(\cfrac{y}{x})=\log x+C$$
Question 7
1 / -0

What is the solution of $$(1+2x)dy-(1-2y)dx=0$$?

A
$$x-y-2xy=c$$

B
$$y-x-2xy=c$$

C
$$y+x-2xy=c$$

D
$$x+y+2xy=c$$

Solution

Consider $$(1+2x)dy-(1-2y)dx=0$$

$$\Rightarrow (1+2x)dy=(1-2y)dx$$

$$\Rightarrow \dfrac{dx}{1+2x}=\dfrac{dy}{1-2y}$$

$$\Rightarrow \int \dfrac{dx}{1+2x}=\int \dfrac{dy}{1-2y}$$

$$\Rightarrow \dfrac{1}{2}log(1+2x)=-\dfrac{1}{2}log(1-2y)+\dfrac{1}{2}log\:c$$

$$\Rightarrow log(1+2x)+log(1-2y)=log\:c$$

$$\Rightarrow (1+2x)(1-2y)=c$$

$$\Rightarrow x-y-2xy=c$$
Question 8
1 / -0

Let $$f(x)$$ be a differential function and satisfy $$f(0) =2$$, and $$f'(x) = f(x)$$ . Find

A
The range of the function $$f(x)\ is\ (0,\infty)$$

B
The value of the function when $$x= ln2\ is\ 3$$

C
The area enclosed by $$y =f(x)$$ in the $$2nd$$ quadrant is bounded.

D
None of these

Solution

$$f^{'}(x)=f(x)$$
$$\Rightarrow \cfrac{dy}{dx}=y$$
$$\Rightarrow\cfrac{dy}{y}=dx$$
Now on integrating, we get
$$\log y=x+C$$
$$f(0)=2\Rightarrow C=\log 2$$
$$\log y=x+C=x+\log2$$
$$y=2e^{x}$$
Therefore, we can say that range of f(x) is $$(0,\infty )$$.
Question 9
1 / -0

The solution of $$\left( {\cos ecx\log y} \right)dy + \left( {{x^2}y} \right)dx = 0$$ is

A
$${{\log y} \over 2} + \left( {2 - {x^2}} \right)\cos x + 2\sin x = c$$

B
$${\left( {{{\log y} \over 2}} \right)^2} + \left( {2 - {x^2}} \right)\cos x + 2x\sin x = c$$

C
$${{{{\left( {\log y} \right)}^2}} \over 2} + \left( {2 - {x^2}} \right)\cos x + 2x\sin x = c$$

D
none of these

Solution

$$\left( {\mathrm cos ec \ x\log y} \right)dy + \left( {{x^2}y} \right)dx = 0$$
$$\cfrac{-\log y}{y}\,dy=\cfrac{x^{2}dx}{\csc x}=\sin x\, x^{2}dx$$
Integrating on both the sides and using Product Formula in RHS; for LHS,
Let $$\log y=z$$
$$\cfrac{1}{y}dy=dz$$
$$\int -z\, dz=x^{2}\int\sin x-\int(\cfrac{d}{dx}x^{2}\int \sin x)dx$$
$$\cfrac{-z^{2}}{2}=-x^{2}\,\cos x+\int\,2x\,\cos x$$
$$\cfrac{-z^{2}}{2}=-x^{2}\,\cos x+2[x\int \cos x-\int(\cfrac{dx}{dx}\int\cos x) dx]$$
$$\cfrac{-z^{2}}{2}=-x^{2}\,\cos x+2x\sin x-2\int \sin x$$
$$\cfrac{(-\log y^{2})}{2}=-x^{2}\,\cos x+2x\sin x+2 \cos x+C$$
$$(2-x^{2})\cos x+\cfrac{(log\, y)^{2}}{2}+2x\sin x=C$$
Question 10
1 / -0

What is the solution of the differential equation $$xdy+ydx=0$$

A
$$xy=c$$

B
$$y=cx$$

C
$$x+y=c$$

D
$$x-y=c$$

Solution

Consider $$xdy+ydx=0$$

$$\Rightarrow xdy=-ydx$$

$$\Rightarrow \dfrac{1}{y}dy=-\dfrac{1}{x}dx$$

Integrating both sides we get

$$\Rightarrow ln\:|y|=-ln\:|x|+C$$

$$\Rightarrow ln\:|y|+ln\:|x|=C$$

$$\Rightarrow ln(|x||y|)=C$$

$$\Rightarrow |x||y|=e^C$$

$$\Rightarrow xy=e^C$$

Let $$e^C=c$$ we get

$$\Rightarrow xy=c$$

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Differential Equations Test - 53

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Differential Equations Test - 53

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SHARING IS CARING

Question 1

$$\frac{1}{y(x)} \int y_1(x) dx$$

$$y_1(x) \int \frac{r(x)}{y_1(x)} dx$$

$$ \int r(x)y_1(x) dx$$

none of these

Solution

Question 2

$$log|1+tan \frac{x+y}{2}| = y+c$$

$$log|2+sec \frac{x+y}{2}| = x+c$$

$$log|1+tan (x+y)| = y+c$$

none of these

Solution

Question 3

$$log |1 + \tan\frac{x+y}{2} | = y+c$$

$$log |1 + \sec\frac{x+y}{2} | = x+c$$

$$log |1 + \tan(x+y) | = y+c$$

none of these

Solution

Question 4

$$a = b = 0$$

$$a = -b\neq 0$$

$$a = b \neq 0, h = k$$

$$a = b \neq 0$$

Solution

Question 5

$$({ y-1 }^{ )2 }{ (1- }{ x }^{ 2 })=0$$

$$({ y-1 }^{ )2 }{ (1- }{ x }^{ 2 })={ c }^{ 2 }{ y }^{ 2 }$$

$$({ y-1 }^{ )2 }{ (1+ }{ x }^{ 2 })={ c }^{ 2 }{ y }^{ 2 }$$

None of these

Solution

Question 6

$$tan^{-1}\dfrac{y}{x} = log x + c$$

$$tan^{-1}\dfrac{y}{x} = -log x + c$$

$$sin^{-1}\dfrac{y}{x} = log x + c$$

$$tan^{-1}\dfrac{x}{y} = log x + c$$

Solution

Question 7

$$x-y-2xy=c$$

$$y-x-2xy=c$$

$$y+x-2xy=c$$

$$x+y+2xy=c$$

Solution

Question 8

The range of the function $$f(x)\ is\ (0,\infty)$$

The value of the function when $$x= ln2\ is\ 3$$

The area enclosed by $$y =f(x)$$ in the $$2nd$$ quadrant is bounded.

None of these

Solution

Question 9

$${{\log y} \over 2} + \left( {2 - {x^2}} \right)\cos x + 2\sin x = c$$

$${\left( {{{\log y} \over 2}} \right)^2} + \left( {2 - {x^2}} \right)\cos x + 2x\sin x = c$$

$${{{{\left( {\log y} \right)}^2}} \over 2} + \left( {2 - {x^2}} \right)\cos x + 2x\sin x = c$$

none of these

Solution

Question 10

$$xy=c$$

$$y=cx$$

$$x+y=c$$

$$x-y=c$$

Solution

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