第12类NCERT解决方案-数学第二部分–第9章微分方程-练习-9.3

在练习1到5中的每一个中，通过消除任意约束a和b，从表示给定曲线族的微分方程中获得。

问题1 $\frac{x}{a}+\frac{y}{b}=1$

解决方案：

Given: $\frac{x}{a}+\frac{y}{b}=1$

We can also write

 bx + ay = ab

On differentiating we get

b + ay’ = 0

y’ = -b/a

Again differentiating we get

y” = 0               

问题2。 $y^2=a(b^2-x^2)$

解决方案：

Given: $y^2=a(b^2-x^2)$

On differentiating we get

2y.y’=-2ax

$\frac{yy'}{x}=-a$

Again differentiating we get

$\frac{x[yy''+(y')2]-yy'}{x^2}=0$

xyy” + x(y’)² – yy’ = 0   

问题3. y = ae ^3x + be ^-2x

解决方案：

 y = ae^3x + be^-2x         -(1)

On differentiating we get

y’=3ae^3x-2be^-2x         -(2)

Again differentiating we get

y”=9ae^3x+4be^2x       

Now on multiply eq(1) by 6 

6y = 6ae^3x + 6be^-2x

And add with eq(2) 

6y + y’ = 6ae^3x + 6be^-2x + 3ae^3x – 3be^-2x

6y + y’ = 9ae^3x + 4be^-2z = y”

y” – y’ – 6y = 0  

问题4. y = e ^2x (a + bx)

解决方案：

Given: y = e^2x(a + bx)         -(1)

On differentiating we get

y’ = e^2x(b) + (a + bx).2e^2x

y’ = e^2x(b + 2a + 2bx)        -(2)

Now on multiply eq(1) by 2

2y = e^2x(2a + 2bx)

And add with eq(2) 

y’ – 2y = e^2x(b + 2a + 2bx) – e^2x(2a + 2bx)

y’ – 2y = be^2x        -(3)

Again differentiating we get

y” – 2y’ = 2be^2x

Now put the value of be^2x from eq(3)

y” – 2y’ = 2(y’ – 2y)

y” – 2y’ = 2y’ – 4y

y” – 2y’ – 2y’ + 4y = 0

y” – 4y’ + 4y = 0

问题5 $y=e^x(a\cos x+b\sin x)$

解决方案：

Given: y = e^x(a cos x + b sin x)         -(1)

On differentiating we get

y’ = e^x[a cos x + b sin x – a sin x + b cos x]

y’ = y + e^x[b cos x – a sin x]          -(2)

Again differentiating we get

y’ ‘ =y’ + e^x[b cos x – a sin x – b sin x – a cos x]

y” = y’ + e^x[b cos x – a sin x] – e^x[a cos x + b sin x]

From eq(1) and (2), we get

y” = y’ + [y’ – y] – y                   

y” – 2y’ + 2y = 0 

问题6：形成在原点接触y轴的圆族的微分方程。

解决方案：

Given that the family of circles touching the y-axis at the origin.

So, the center of the circle is (a, 0) and radius a

Let the equation of a circle is

(x – a)² + y² = a²

= x² + y² = 2ax         -(1)

On differentiating we get

2x + 2yy’ = 2a

x + yy’ = a

Now substitute the value of a in eq(1), we get

x² + y² = 2(x + yy’)x

x² + y² = 2x² + 2xyy’

x² + y² – 2x² – 2xyy’

y² = x² + 2xyy’

问题7.形成抛物线族的微分方程，该抛物线族在原点和沿y轴的正方向具有顶点。

解决方案：

Given that the family of parabolas having a vertex at origin and axis along positive y-axis. 

So the equation of parabola is:

x² = 4ay           -(1)

On differentiating we get

2x = 4ay’           -(2)

Now divide eq(2) by (1), we have 

2x/ x² = 4ay’ /4ay 

2/x = y’ /y 

y’x = 2y 

 y’x – 2y = 0

问题8.形成焦点在y轴且以原点为中心的椭圆族的微分方程。

解决方案：

Given that the family of ellipses having facii or y-axis & center at the origin.

So the equation of parabola is 

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

On differentiating we get

$\frac{2x}{b^2}+\frac{2y}{a^2}y'=0$

$y'=\frac{-xb^2}{ya^2}$

$\frac{yy'}{x}=\frac{-b^2}{a^2}$

Again differentiating we get

$\frac{x[yy''+(y')^2]-yy'}{x^2}=0$

xyy” + x(y’)² – yy’ = 0  

问题9.形成双曲线族的微分方程，该双曲线族的焦点在x轴上，并且以原点为中心。

解决方案：

Given that the family of hyperbolas having foci on the x-axis and center at the origin.

So the equation of hyperbola is 

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

On differentiating we get

$\frac{2x}{a^2}+\frac{2y}{b^2}y'=0$

$y'=\frac{b^2}{a^2y}$

$\frac{yy'}{x}=\frac{b^2}{a^2}$

Again differentiating we get

$\frac{x[yy''+(y')^2]-yy'}{x^2}=0$

$y'\sqrt{9-x^2}-x=0$ 

问题10：形成一个以y轴为中心，半径为3个单位的圆族的微分方程。

解决方案：

Given that the of circles having a center on y-axis and radius 3 units.

So the center be (0, k) 

General equation of the circle is,

x² + (y – k)² = 3²             

(y – k)² = 9 – x²

y – k = $\sqrt{9-x^2}$

k = $y-\sqrt{9-x^2}$

On differentiating we get

$y'-\frac{1}{2\sqrt{9-x^2}}.(-2x)=0$

$y'+\frac{x}{\sqrt{9-x^2}}=0$

Squaring on both sides we get

$(9-x^2)(y')^2+x^2=0$

问题11.以下哪个微分方程具有y = c ₁ e ^x + c ₂ e ^-x作为一般解?

(一种) $\frac{d^2y}{dx^2}+y=0$

(B) $\frac{d^2y}{dx^2}-y=0$

(C) $\frac{d^2y}{dx^2}+1=0$

(D) $\frac{d^2y}{dx^2}-1=0$

解决方案：

y = c₁e^x + c₂e^-x

On differentiating we get

y’ = c₁e^x – c₂e^-x

Again differentiating we get

y” = c₁e^x + c₂e^-x

y” = y

y” – y = 0 

Hence, the correct option is B

问题12.以下哪个微分方程具有y = x作为其特定解之一?

(一种) $\frac{d^2y}{dx^2}-x^2\frac{dy}{dx}+xy=x$

(B) $\frac{d^2y}{dx^2}+x\frac{dy}{dx}+xy=x$

(C) $\frac{d^2y}{dx^2}-x^2\frac{dy}{dx}+xy=0$

(D) $\frac{d^2y}{dx^2}+x\frac{dy}{dx}+xy=0$

解决方案：

y = x

On differentiating we get

y’ = 1

Again differentiating we get

y” = 0

Now substitute the value of y, y’ and y” in each option to check for correct option

(A) $\frac{d^2y}{dx^2}-x^2\frac{dy}{dx}+xy=x$

= 0 – x²(1) + x.x = x

0 ≠ x

(B) $\frac{d^2y}{dx^2}+x\frac{dy}{dx}+xy=x$

= 0 + x(1) + x.x = x

= x + x²≠ x

(C) $\frac{d^2y}{dx^2}-x^2\frac{dy}{dx}+xy=0$

= 0 – x²(1) + x.x = 0

= 0 = 0

Hence, the correct option is C.